Objectives ndash Chapter 15bull Define the Common Ion Effect (151)bull Define buffer and show how a buffer
controls pH of a solution (152 153)bull Identify and Evaluate titration curves
(154)bull Use of Indicators (155)
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Stomach Acidity ampAcid-Base Reactions
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Acid-Base Reactions
bull Strong acid + strong base HCl + NaOH bull Strong acid + weak base
HCl + NH3 bull Weak acid + strong base HOAc + NaOH bull Weak acid + weak base
HOAc + NH3
What is relative pH
before during amp
after reaction
Need to study
a) Common ion
effect and buffers
b) Titrations
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Here we are adding NH4+ an ion COMMON to the
equilibrium
Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)
The pH will go up (1) down (2) no change (3)
NH4+ is an acid
The Common Ion Effect
Section 151
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Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
2
copy 2009 BrooksCole - Cengage
Objectives ndash Chapter 15bull Define the Common Ion Effect (151)bull Define buffer and show how a buffer
controls pH of a solution (152 153)bull Identify and Evaluate titration curves
(154)bull Use of Indicators (155)
3
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Stomach Acidity ampAcid-Base Reactions
4
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Acid-Base Reactions
bull Strong acid + strong base HCl + NaOH bull Strong acid + weak base
HCl + NH3 bull Weak acid + strong base HOAc + NaOH bull Weak acid + weak base
HOAc + NH3
What is relative pH
before during amp
after reaction
Need to study
a) Common ion
effect and buffers
b) Titrations
5
copy 2009 BrooksCole - Cengage
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Here we are adding NH4+ an ion COMMON to the
equilibrium
Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)
The pH will go up (1) down (2) no change (3)
NH4+ is an acid
The Common Ion Effect
Section 151
6
copy 2009 BrooksCole - Cengage
Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
7
copy 2009 BrooksCole - Cengage
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
8
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
9
copy 2009 BrooksCole - Cengage
Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
10
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
11
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
90
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
92
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
98
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
3
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Stomach Acidity ampAcid-Base Reactions
4
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Acid-Base Reactions
bull Strong acid + strong base HCl + NaOH bull Strong acid + weak base
HCl + NH3 bull Weak acid + strong base HOAc + NaOH bull Weak acid + weak base
HOAc + NH3
What is relative pH
before during amp
after reaction
Need to study
a) Common ion
effect and buffers
b) Titrations
5
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Here we are adding NH4+ an ion COMMON to the
equilibrium
Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)
The pH will go up (1) down (2) no change (3)
NH4+ is an acid
The Common Ion Effect
Section 151
6
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Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
8
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
9
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
10
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
11
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
13
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
14
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
copy 2009 BrooksCole - Cengage
Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
107
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Acid-Base Reactions
bull Strong acid + strong base HCl + NaOH bull Strong acid + weak base
HCl + NH3 bull Weak acid + strong base HOAc + NaOH bull Weak acid + weak base
HOAc + NH3
What is relative pH
before during amp
after reaction
Need to study
a) Common ion
effect and buffers
b) Titrations
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Here we are adding NH4+ an ion COMMON to the
equilibrium
Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)
The pH will go up (1) down (2) no change (3)
NH4+ is an acid
The Common Ion Effect
Section 151
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Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
10
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
11
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Here we are adding NH4+ an ion COMMON to the
equilibrium
Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)
The pH will go up (1) down (2) no change (3)
NH4+ is an acid
The Common Ion Effect
Section 151
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Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
copy 2009 BrooksCole - Cengage
Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Let us first calculate the pH of a 025 M NH3 solution
[NH3] [NH4+] [OH-]
initial 025 0 0change -x +x +xequilib 025 - x x x
QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NH3
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
copy 2009 BrooksCole - Cengage
Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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copy 2009 BrooksCole - Cengage
Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
copy 2009 BrooksCole - Cengage
Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
025 - x
pH of Aqueous NH3
Assuming x is ltlt 025 we have
[OH-] = x = [Kb(025)]12 = 00021 M
This gives pOH = 267
and so pH = 1400 - 267
= 1133 for 025 M NH3
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
30
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initialchangeequilib
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
9
copy 2009 BrooksCole - Cengage
Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding NH4Cl Letrsquos test that
[NH3] [NH4+] [OH-]
initial 025 010 0change -x +x +xequilib 025 - x 010 + x x
pH of NH3NH4+ Mixture
10
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
90
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
92
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
Kb= 18 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(010 + x)
025 - x
pH of NH3NH4+ Mixture
Assuming x is very small
[OH-] = x = (025 010)(Kb) = 45 x 10-5 M
This gives pOH = 435 and pH = 965pH drops from 1133 to 965
on adding a common ion
11
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
12
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
14
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
15
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
30
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Common Ion Effectbull The addition of a common ion in a base such as ammonia
limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)
B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit
ionization of acid which limits [H3O+]
HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid
because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl which has a very large Ka
HCl + H2O H3O+ + Cl-
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Common Ion Effect Practice
Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+
With salt
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
14
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Buffered SolutionsSection 152
HCl is added to pure water
HCl is added to a solution of a weak acid H2PO4
- and its conjugate base HPO4
2-
15
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
30
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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A buffer solution is a special case of the common ion effect
The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Consider HOAcOAc- to see how buffers workACID USES UP ADDED OH-
We know that
OAc- + H2O HOAc + OH-
has Kb = 56 x 10-10
Therefore the reverse reaction of the WEAK ACID with added OH-
has Kreverse = 1 Kb = 18 x 109
Kreverse is VERY LARGE so HOAc completely consumes OH-
Buffer Solutions
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Consider HOAcOAc- to see how buffers work
CONJ BASE USES UP ADDED H+
HOAc + H2O OAc- + H3O+
has Ka = 18 x 10-5
Therefore the reverse reaction of the WEAK BASE with added H+
has Kreverse = 1 Ka = 56 x 104
Kreverse is VERY LARGE so OAc- completely consumes H3O+
Buffer Solutions
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Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
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[HOAc] [OAc-] [H3O+]
initialchangeequilib
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 0600 0-x +x +x
0700 - x 0600 + x x
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[HOAc] [OAc-] [H3O+]
equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have
Ka= 18 x 10-5 =
[H3O+ ](0600)
0700
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
[H3O+] = 21 x 10-5 and pH = 468
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Consider HOAcOAc- to see how buffers work
CONJ BASE USES UP ADDED H+
HOAc + H2O OAc- + H3O+
has Ka = 18 x 10-5
Therefore the reverse reaction of the WEAK BASE with added H+
has Kreverse = 1 Ka = 56 x 104
Kreverse is VERY LARGE so OAc- completely consumes H3O+
Buffer Solutions
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Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
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[HOAc] [OAc-] [H3O+]
initialchangeequilib
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 0600 0-x +x +x
0700 - x 0600 + x x
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[HOAc] [OAc-] [H3O+]
equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have
Ka= 18 x 10-5 =
[H3O+ ](0600)
0700
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
[H3O+] = 21 x 10-5 and pH = 468
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
0700 M HOAc has pH = 245
The pH of the buffer will have
1 pH lt 245
2 pH gt 245
3 pH = 245
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[HOAc] [OAc-] [H3O+]
initialchangeequilib
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 0600 0-x +x +x
0700 - x 0600 + x x
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[HOAc] [OAc-] [H3O+]
equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have
Ka= 18 x 10-5 =
[H3O+ ](0600)
0700
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
[H3O+] = 21 x 10-5 and pH = 468
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
22
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
copy 2009 BrooksCole - Cengage
Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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[HOAc] [OAc-] [H3O+]
initialchangeequilib
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
0700 0600 0-x +x +x
0700 - x 0600 + x x
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[HOAc] [OAc-] [H3O+]
equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have
Ka= 18 x 10-5 =
[H3O+ ](0600)
0700
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
[H3O+] = 21 x 10-5 and pH = 468
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
23
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
37
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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[HOAc] [OAc-] [H3O+]
equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have
Ka= 18 x 10-5 =
[H3O+ ](0600)
0700
Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M
HOAc + H2O OAc- + H3O+
Ka = 18 x 10-5
Buffer Solutions
[H3O+] = 21 x 10-5 and pH = 468
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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copy 2009 BrooksCole - Cengage
Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Notice that the expression for calculating the H+ conc of the buffer is
[H3O+ ] = Orig conc of HOAc
Orig conc of OAc- x Ka
Buffer Solutions
Notice that the [H3O+] depends on
1the Ka of the acid2the ratio of the [acid] and
[conjugate base]
[H3O+ ] =
[Acid]
[Conj base] x Ka
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Henderson-Hasselbalch Equation
Take the negative log of both sides of this equation
The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid
or
Note HA and A- are generic terms for acid and conjugate base respectively
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Henderson-Hasselbalch EquationWorks the same for base
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Solve for [OH-] =
Take the negative log of both sides of this equation to get
The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer Calculation Example(AcidConj Base)
What is the pH of a buffer that is 012 M in lactic acid HC3H5O3 and 010 M in sodium lactate (NaC3H5O3)
Ka for lactic acid is 14 10-4
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Buffer Calculation Example
HC3H5O3 + H2O C3H5O3- + H3O+
I 012 M 010 - E 012-x 010 + x x
x(010 + x) = 14x10-4
012 - xWersquoll make usual assumption that x is small
x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]
010 So the pH = 377
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Buffer Calculation using the HendersonndashHasselbalch Equation
pH = ndashlog (14 10-4) + log(010)(012)
pH = 385 + (ndash008)
pH = 377 (same as before)
HC3H5O3 + H2O C3H5O3- + H3O+
acid conj base 012 010
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Buffer Calculation using the HendersonndashHasselbalch Equation
pOH = ndashlog (18 x 10-5) + log(0045)(0080)
Calculate the pH of a buffer that has is 0080 M ammonia and 0045 M ammonium chlorideThe Kb for ammonia is 18 x 10-5 NH3(aq) + H2O(l) NH4
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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copy 2009 BrooksCole - Cengage
Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer Calculation Example
HC3H5O3 + H2O C3H5O3- + H3O+
I 012 M 010 - E 012-x 010 + x x
x(010 + x) = 14x10-4
012 - xWersquoll make usual assumption that x is small
x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]
010 So the pH = 377
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Buffer Calculation using the HendersonndashHasselbalch Equation
pH = ndashlog (14 10-4) + log(010)(012)
pH = 385 + (ndash008)
pH = 377 (same as before)
HC3H5O3 + H2O C3H5O3- + H3O+
acid conj base 012 010
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Buffer Calculation using the HendersonndashHasselbalch Equation
pOH = ndashlog (18 x 10-5) + log(0045)(0080)
Calculate the pH of a buffer that has is 0080 M ammonia and 0045 M ammonium chlorideThe Kb for ammonia is 18 x 10-5 NH3(aq) + H2O(l) NH4
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer Calculation using the HendersonndashHasselbalch Equation
pH = ndashlog (14 10-4) + log(010)(012)
pH = 385 + (ndash008)
pH = 377 (same as before)
HC3H5O3 + H2O C3H5O3- + H3O+
acid conj base 012 010
27
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Buffer Calculation using the HendersonndashHasselbalch Equation
pOH = ndashlog (18 x 10-5) + log(0045)(0080)
Calculate the pH of a buffer that has is 0080 M ammonia and 0045 M ammonium chlorideThe Kb for ammonia is 18 x 10-5 NH3(aq) + H2O(l) NH4
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
37
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
90
copy 2009 BrooksCole - Cengage
Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
98
copy 2009 BrooksCole - Cengage
Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
27
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Buffer Calculation using the HendersonndashHasselbalch Equation
pOH = ndashlog (18 x 10-5) + log(0045)(0080)
Calculate the pH of a buffer that has is 0080 M ammonia and 0045 M ammonium chlorideThe Kb for ammonia is 18 x 10-5 NH3(aq) + H2O(l) NH4
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
29
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
30
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
37
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer Practice 1
Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2
(Ka = 18 x 10-5) Use H-H Eqn
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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copy 2009 BrooksCole - Cengage
Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400
(Benzoic acid Ka = 63 x 10-5)
Use Henderson-Hasselbalch equation
HC7H5O2 + H2O H3O+ + C7H5O2-
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
30
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl
is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 999 x 10-4 M = [H3O+]
pH = 300
31
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
32
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Adding an Acid to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometry
The reaction occurs completely because K is very large
The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
33
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Adding an Acid to a Buffer
Solution to Part (b) Step 1mdashStoichiometry
[H3O+] + [OAc-] [HOAc]
Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
000100 mol0600 mol 0700 mol
-000100 -000100 +000100
0 0599 mol 0701 mol
0 0599 mol1001 L
0701 mol 100l L
0598 molL0700 molL
34
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
35
copy 2009 BrooksCole - Cengage
Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial (M)Change (M)Equil (M) (re-established)
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
0700 molL0 0598 molL-x +x +x
0598 + xx0700-x
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
35
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0700-x x 0598+x
Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
36
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Adding an Acid to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468)
pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer
37
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
38
copy 2009 BrooksCole - Cengage
Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
copy 2009 BrooksCole - Cengage
Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
44
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
37
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Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M
NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of
water
M1bullV1 = M2 bull V2
(100 M)(10 mL) = M2(1001 mL)
M2 = 100 x 10-3 M = [OH-]
pOH = 300 pH = 1100
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Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
41
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
38
copy 2009 BrooksCole - Cengage
Adding an Base to a Buffer
Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr
OAc- (from buffer and rxn)The reaction occurs completely because K is
very largeThe strong base (OH-) will react with weak acid
(HOAc) to make the conj weak base (OAc-)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and
[OAc-] = 0600 M (pH before = 468)
39
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
40
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding a Base to a Buffer
Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
000100 mol0700 mol 0600 mol
-000100 -000100 +000100
0 0699 mol 0601 mol0 0699 mol
1001 L0601 mol 1001 L
0698 molL0600 molL
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
42
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Initial(M)Change (M)Equil (M)(re-established)
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
0698 molL0 0600 molL-x +x +x
0600 + xx0698-x
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [H3O+] [OAc-]
Equilibrium 0698-x x 0600+x
Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698
What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468 pOH = 932)
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Adding a Base to a Buffer
Solution to Part (b) Step 2mdashEquilibrium
HOAc + H2O H3O+ + OAc-
What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]
= 0600 M (pH = 468)
pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
43
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Preparing a BufferYou want to buffer a solution at pH = 430
This means [H3O+] = 10-pH =10-430= 50 x 10-5 M
It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)
mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base
[H3O+ ] =
[Acid]
[Conj base] x Ka
44
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
85
copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
44
copy 2009 BrooksCole - Cengage
You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
POSSIBLE
ACIDSCONJ BASE PAIRS Ka
HSO4- SO4
2- 12 x 10-2
HOAc OAc- 18 x 10-5
HCN CN- 40 x 10-10
Best choice is acetic acid acetate
Preparing a Buffer
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
45
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You want to buffer a solution at pH = 430 or
[H3O+] = 50 x 10-5 M
[H3O+ ] = 50 x 10-5 = [HOAc]
[OAc- ] (18 x 10-5 )
Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430
Solve for [HOAc][OAc-] ratio =
278
1
Preparing a Buffer
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
88
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
92
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
46
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A final point mdashCONCENTRATION of the acid and conjugate
base are not as important as the RATIO OF THE NUMBER OF MOLES of
each
Result diluting a buffer solution does not change its pH
Preparing a Buffer
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
49
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
47
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Commercial Buffers
bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH
bull Note that the quantity of water does not affect the pH of the buffer
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
48
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Buffer prepared from
84 g NaHCO3 (84g84 gmol)
weak acid
160 g Na2CO3 (160106 gmol)
conjugate base
HCO3- + H2O
H3O+ + CO32-
What is the pH
HCO3- pKa = 103
pH = 103 + log (15110)=105
Preparing a Buffer
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Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
85
copy 2009 BrooksCole - Cengage
OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
49
copy 2009 BrooksCole - Cengage
Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base
In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 721 what should the
[HPO42-][H2PO4
-] ratio be
B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer
(assume the Na2HPO4 doesnrsquot affect the volume)
50
copy 2009 BrooksCole - Cengage
Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
copy 2009 BrooksCole - Cengage
Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
copy 2009 BrooksCole - Cengage
Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Buffer ExamplePart A
H2PO4-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Can use H-H pH = pKa + log[HPO42-][H2PO4
-]
740 = 721 + log[HPO42-][H2PO4
-]
log[HPO42-][H2PO4
-] = 019
[HPO42-][H2PO4
-] = 10019 = 155
Part B
60g NaH2PO4 = 0050 mol in 0500 L = 010 M
120gmol
[HPO42-] = 155[H2PO4
-] = 0155 M HPO42-
=0155 molL 0500 L 142 g Na2HPO4mol
= 110 g Na2HPO4
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
51
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Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)
52
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Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
52
copy 2009 BrooksCole - Cengage
Buffering CapacityChapter 153
bull The pH of a buffered solution is
determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much
the pH wonrsquot change much
bull The more concentrated these two are the
more H+ and OH- the solution will be able to absorb
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
53
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Buffer Capacity
bull Calculate the change in pH that occurs when 0010 mol of HCl is added to 10L of each of the following
bull 500 M HOAc and 500 M NaOAcbull 0050 M HOAc and 0050 M NaOAc
bull Ka= 18x10-5 (pKa = 474)bull Note Since [OAc-]=[HOAc] for both cases
the initial pH is 474
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Large Buffer System
H3O+ + OAc- HOAc + H2O
H3O+ OAc- HOAc
Before Reaction (moles) 001 500 500
After Reaction (moles) 0 499 501
Stoichiometry Calculation
55
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
0501 X 499 [base]pH = pKa +
[acid]
pH = -log (18 x 10-5) + log[ 499501]
pH = 4738
Since pKa = 474 solution has gotten slightly more acidic
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Small Buffer System
H3O+ + OAc- HOAc(aq) + H2O
H+ OAc- HOAc
Before Reaction (moles) 001 005 005
After Reaction (moles) 0 004 006
Stoichiometry Calculation
57
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Calculating pH for Small Buffer System
In table form (Equilibrium Re-established)
[HOAc] M [H3O+] M [OAcminus] M
Initially 006 0 004
Change -x +x +x
At Equilibrium
006-x x 004 + x
HOAc H3O+ + OAc-
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Large Buffer System
H3O+ + OAc- HOAc + H2O
H3O+ OAc- HOAc
Before Reaction (moles) 001 500 500
After Reaction (moles) 0 499 501
Stoichiometry Calculation
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
0501 X 499 [base]pH = pKa +
[acid]
pH = -log (18 x 10-5) + log[ 499501]
pH = 4738
Since pKa = 474 solution has gotten slightly more acidic
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Small Buffer System
H3O+ + OAc- HOAc(aq) + H2O
H+ OAc- HOAc
Before Reaction (moles) 001 005 005
After Reaction (moles) 0 004 006
Stoichiometry Calculation
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Calculating pH for Small Buffer System
In table form (Equilibrium Re-established)
[HOAc] M [H3O+] M [OAcminus] M
Initially 006 0 004
Change -x +x +x
At Equilibrium
006-x x 004 + x
HOAc H3O+ + OAc-
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
55
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
0501 X 499 [base]pH = pKa +
[acid]
pH = -log (18 x 10-5) + log[ 499501]
pH = 4738
Since pKa = 474 solution has gotten slightly more acidic
56
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Small Buffer System
H3O+ + OAc- HOAc(aq) + H2O
H+ OAc- HOAc
Before Reaction (moles) 001 005 005
After Reaction (moles) 0 004 006
Stoichiometry Calculation
57
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Calculating pH for Small Buffer System
In table form (Equilibrium Re-established)
[HOAc] M [H3O+] M [OAcminus] M
Initially 006 0 004
Change -x +x +x
At Equilibrium
006-x x 004 + x
HOAc H3O+ + OAc-
58
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
56
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Small Buffer System
H3O+ + OAc- HOAc(aq) + H2O
H+ OAc- HOAc
Before Reaction (moles) 001 005 005
After Reaction (moles) 0 004 006
Stoichiometry Calculation
57
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Calculating pH for Small Buffer System
In table form (Equilibrium Re-established)
[HOAc] M [H3O+] M [OAcminus] M
Initially 006 0 004
Change -x +x +x
At Equilibrium
006-x x 004 + x
HOAc H3O+ + OAc-
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
88
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
92
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
117
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
57
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Calculating pH for Small Buffer System
In table form (Equilibrium Re-established)
[HOAc] M [H3O+] M [OAcminus] M
Initially 006 0 004
Change -x +x +x
At Equilibrium
006-x x 004 + x
HOAc H3O+ + OAc-
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
58
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Calculate pH via Henderson-Hasselbalch Equations
At Equilibrium HOAc + H2O H3O+ + OAc-
006-x X 004-x
Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn
So solution is considerably more acidic than the pH 474 of the big buffer system
59
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Chapter 154
If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically
H+ + OH- H2O(l)
I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
60
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
85
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
88
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
89
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
95
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
99
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
61
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH
H+ + OH- H2O(l)
I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-
pOH = 177 so pH = 1223
62
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
HOAc + OH- H2O(l) + OAc-
I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
63
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
64
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established
H2O(l) + OAc- HOAc + OH-
E 0050-x x x
Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10
= 556x10-10
Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
65
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
66
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Mixing Acids and Bases(Stoichiometry amp
Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH CurvesTitrations and pH Curves
pHpH
Titrant volume mLTitrant volume mL
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Titrations and pH Curves
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very
slowly
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
69
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Titrations and pH Curves
Additional NaOH is added pH rises as equivalence point is approached
>
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
70
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Titrations and pH Curves
Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence
point
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
85
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
88
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
71
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
What is the pH at equivalence point
What is the pH at equivalence point
pH of solution of benzoic acid a weak acid
pH of solution of benzoic acid a weak acidBenzoic acid
+ NaOHBenzoic acid + NaOH
What is pH at half-way point
What is pH at half-way point
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
82
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
88
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
72
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Titrations and pH Curves
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
C6H5CO2H = HBz Benzoate ion = Bz-
73
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
74
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
75
copy 2009 BrooksCole - Cengage
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
76
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
77
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
78
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
79
copy 2009 BrooksCole - Cengage
Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
80
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
HBz + NaOH Na+ + Bz- + H2O
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
The product of the titration of benzoic acid is the benzoate ion Bz-
Bz- is the conjugate base of a weak acid Therefore solution is basic at
equivalence point
Bz- + H2O HBz + OH-
Kb = 16 x 10-10
++
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point
pH at equivalence point is basic
pH at equivalence point is basic
Benzoic acid + NaOHBenzoic acid + NaOH
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH
This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
77
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Titrations and pH Curves
STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
78
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Titrations and pH Curves
STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =
00025 mol4 Calc conc of Bz-
There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point
125 mL
[Bz-] = 00025 mol 0125 L = 0020 M
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
79
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate
ion Bz- the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
Equivalence PointMost important species in solution is benzoate ion Bz-
the weak conjugate base of benzoic acid HBz
Bz- + H2O HBz + OH- Kb = 16 x 10-10
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point
x = [OH-] = 18 x 10-6
pOH = 575 and pH = 825
Kb = 16 x 10-10 =
x2
0020 - x
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
104
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
81
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point1 lt 72 = 73 gt 7
pH at half-way point1 lt 72 = 73 gt 7
Equivalence point pH = 825
Equivalence point pH = 825
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
83
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
84
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
107
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point
pH at half-way point
pH at half-way point Equivalence point
pH = 825
Equivalence point pH = 825
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What is the pH at the half-way point
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5
pH = 420 = pKa of the acid
Both HBz and Bz- are present
This is a BUFFER
Both HBz and Bz- are present
This is a BUFFER
HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
What was the pH when 60 mL of NaOH was added
StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution
When doing these types of problems start from initial moles and go to the point you are considering
OH- + HBz H2O + Bz-
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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OH- + HBz H2O + Bz-
Titrations and pH Curves
You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH
Practice What is the pH when 200 mL of NaOH are added
EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Acetic acid titrated with NaOH
Weak acid titrated with a strong base
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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See Figure 184
Strong acid titrated with a strong base
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Weak diprotic acid (H2C2O4) titrated with a strong base
(NaOH)
See Figure 186
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base
pH
Volume of titrating reagent added --gt
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Weak base (NH3) titrated with a strong
acid (HCl)
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Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
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Acid-Base IndicatorsAcid-Base Indicators
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
91
copy 2009 BrooksCole - Cengage
Acid-Base IndicatorsChapter 155
bull Weak acids that change color when they become bases
bull Since the Indicator is a weak acid it has a Ka
bull End point - when the indicator changes color
bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator
bull Want an indicator where pKa pH at equiv pt
92
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
92
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Acid-Base IndicatorsAcid-Base Indicators
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
95
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
96
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
98
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
99
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
copy 2009 BrooksCole - Cengage
Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
104
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
93
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Indicators for Acid-Base Titrations
Indicators for Acid-Base Titrations
Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
104
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
107
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
94
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Titration of a Base with an Acid
bull The pH at the equivalence point in these titrations is lt 7
bull Methyl red is the indicator of choice
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
95
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Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria
Chapter 16
Lead(II) iodide
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Objectives ndash Chapter 16
bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)
bull Manipulate solubility by common ion effect (161)
bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)
bull Define Complex ion amp show how complex ions affect solubility (163)
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
bull Apply equilibrium principles to acid-base and precipitation reactions
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Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
98
copy 2009 BrooksCole - Cengage
Analysis of Silver Group
Analysis of Silver Group
All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
99
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
copy 2009 BrooksCole - Cengage
Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
99
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Analysis of Silver
Group
Analysis of Silver
GroupAlthough all salts formed in this experiment are
said to be insoluble they do dissolve to some
SLIGHT extent
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established no more
AgCl dissolves and the solution is SATURATED
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
102
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
103
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
104
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
107
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
copy 2009 BrooksCole - Cengage
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
100
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M
This is equivalent to the SOLUBILITY of AgCl
What is [Cl-]
[Cl-] = [Ag+] = 167 x 10-5 M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
101
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Analysis of Silver
Group
Analysis of Silver
GroupAgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 167 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (167 x 10-5)(167 x 10-5)
= 279 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Analysis of Silver
Group
Analysis of Silver
Group
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 279 x 10-10
Because this is the product of ldquosolubilitiesrdquo we call it
Ksp = solubility product
constant
bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL
(Can also be expressed as gL or mgL)bull Solubility Product has no units
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
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Some Values of Ksp
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
103
copy 2009 BrooksCole - Cengage
Solubility ProductsIn general for the equilibrium for the reaction is expressed as
MmAa(s) mM+(aq) + aA-(aq)
The Ksp equilibrium expression is
Ksp =[M+]m[A-]a
The higher the Ksp the more soluble the salt is
104
copy 2009 BrooksCole - Cengage
Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
104
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Some Values of Ksp
105
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
106
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
107
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
117
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
120
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5 = [Pb2+][Clndash]2
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Solution
1 Solubility = [Pb2+] = 130 x 10-3 M
[I-] =
[I-] = 2 x [Pb2+] = 260 x 10-3 M
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
107
copy 2009 BrooksCole - Cengage
Solution
2 Ksp = [Pb2+] [I-]2
= [Pb2+] 2 bull [Pb2+]2
Ksp = 4 [Pb2+]3
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
= 4 (solubility)3= 4 (solubility)3
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 000130 M
108
copy 2009 BrooksCole - Cengage
Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
109
copy 2009 BrooksCole - Cengage
SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
copy 2009 BrooksCole - Cengage
Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
108
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Caveat
3 The value we just calculated by
solubility gives a ball-park value
The actual Ksp of PbI2 is 98 x 10-9
Note Calculating Ksp from solubility
gives approximate values that can be off by a few OOMs due to ionic interactions
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide
Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
109
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SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61
mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)
CaC2O4 (s) Ca2+(aq) + C2O42-(aq)
110
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SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
118
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
110
copy 2009 BrooksCole - Cengage
SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the
Ksp is 39x10-11 (Molar Mass = 781 gmol)
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
112
copy 2009 BrooksCole - Cengage
Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
113
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
111
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The Common Ion EffectAdding an ion ldquocommonrdquo to an
equilibrium causes the equilibrium to shift back to reactant
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
120
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
112
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Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)
Ksp = 19 x 10-5
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Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
113
copy 2009 BrooksCole - Cengage
Barium
Sulfate
Ksp = 11 x 10-10
(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines
(a) BaSO4 is a common mineral appearing a white powder or colorless crystals
114
copy 2009 BrooksCole - Cengage
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
115
copy 2009 BrooksCole - Cengage
SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
116
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
117
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2
Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)12 = 11 x 10-5 M
Solubility in pure water = 11 x 10-5 molL
The Common Ion Effect
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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SolutionSolubility in pure water = 11 x 10-5 molL
Now dissolve BaSO4 in water already containing 0010 M Ba2+
Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
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Solubility and Complex Ions
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Solution
[Ba2+] [SO42-]
initialchange equilib
The Common Ion Effect
+ y0010 0
+ y
0010 + y y
Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
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Separations by Difference in Ksp
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
117
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Solution
Ksp = [Ba2+] [SO42-] = (0010 + y) (y)
Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore
Ksp = 11 x 10-10 = (0010)(y)
y = 11 x 10-8 M = solubility in presence of added Ba2+ ion
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
118
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SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+
= 11 x 10-8 MLe Chatelierrsquos Principle is followed
The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
120
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
119
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Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer
Ksp(AgI) = 85 x 10-17
AgI(s) Ag+(aq) + I-(aq)
120
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
120
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Factors Affecting Solubility - pH
bull pHndash If a substance has a
basic anion (like OH-) it is more soluble in a more acidic solution
ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
121
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Solubility in more Acidic Solution example solid Mg(OH)2 in water
Equilibrium Mg(OH)2 Mg2+ + 2OH-
For Ksp = 18 x 10-11 calculate pH
18 x 10-11 = x(2x)2
x = 165 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 middot 165 x 10-4) = 348pH = 1052
122
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
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Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9
bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp
only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more
acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
123
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Acidic solution example Ag3PO4
bull Ag3(PO4) is more soluble in acid solution than neutral solution Why
ndash H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42-
Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18
H+(aq) + PO43- (aq) HPO4
2-(aq) K = 1Ka = 28x1012
Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6
ndash Addition of H+ shifts equilibrium of 1st rxn to right
ndash PO43- is actually a moderately strong base
PO43- + H2O HPO4
2- + OH- Kb3 = 28x10-2
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
124
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Acidic solution example AgClbull What about AgCl in an acidic solution such
as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10
H+(aq) + Cl- (aq)larr HCl(aq) K = very small
bull K of reaction remains very small largely determined by the Ksp of AgCl
bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less
soluble by Le Chatelier
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
125
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Solubility and Aciditybull Other conjugate base anions that will increase in
solubility in an acidic solution
OH- F- S2- CO32- C2O4
2- and CrO42-
Why Because all form weak acids
Examples
PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
126
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Precipitation and Qualitative AnalysisSection 162
Precipitation and Qualitative AnalysisSection 162
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just
begin the precipitation of Hg2Cl2
That is what is the maximum [Cl-] that can be
in solution with 0010 M Hg22+ without
forming Hg2Cl2
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
127
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs
Precip begins when product of
ion concentrations (Q)
EXCEEDS the Ksp
128
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Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
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AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
128
copy 2009 BrooksCole - Cengage
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0010 M
[Cl- ] =
Ksp
0010 = 11 x 10-8 M
If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in KspSolution
136
copy 2009 BrooksCole - Cengage
Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
129
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Precipitation and Qualitative Analysis
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 11 x 10-18
Now raise [Cl-] to 10 M What is the value of [Hg2
2+] at this point
Solution
[Hg22+] = Ksp [Cl-]2
= Ksp (10)2 = 11 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
130
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Precipitation and Qualitative Analysisndash Practice
bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in KspSolution
136
copy 2009 BrooksCole - Cengage
Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
131
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Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Metal Ions Cu2+ Ag+ Pb2+
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
Ksp Values
AgCl 18 x 10-10
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
132
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Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp
bull Add CrO42- to solid PbCl2 The less
soluble salt PbCrO4 precipitates
bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
bull Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
133
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Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
133
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in Ksp
PbCl2(s) + CrO42- PbCrO4 + 2 Cl-
Salt Ksp
PbCl2 17 x 10-5
PbCrO4 18 x 10-14
PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp
Pb2+ + CrO42- PbCrO4 K2 = 1Ksp
Knet = (K1)(K2) = 94 x 108
Net reaction is product-favored
134
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
copy 2009 BrooksCole - Cengage
Separating Salts by Differences in KspSolution
136
copy 2009 BrooksCole - Cengage
Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
134
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Separating Salts by Differences in Ksp
The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
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Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
135
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Separating Salts by Differences in KspSolution
136
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Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
136
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Separations by Difference in Ksp
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
137
copy 2009 BrooksCole - Cengage
Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions
138
copy 2009 BrooksCole - Cengage
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
138
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The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX
IONS
Equilibria Involving Complex Ions
Chapter 163
139
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Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
139
copy 2009 BrooksCole - Cengage
Reaction of NH3 with Cu2+
(aq)
>
140
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Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
140
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2
+
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
bull The Ag(NH3)2+ is called a complex ion
bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that
(See Chapter 21 for details)
141
copy 2009 BrooksCole - Cengage
Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
141
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Solubility and Complex Ionsbull
bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
142
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Solubility and Complex Ions
143
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Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
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Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
143
copy 2009 BrooksCole - Cengage
Formation of complex ions explains why you can dissolve a ppt by forming a complex ion
Dissolving Precipitates by forming Complex Ions
AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
144
copy 2009 BrooksCole - Cengage
Solubility and Complex Ions
bull Complex Ionsndash The formation
of these complex ions increase the solubility of these salts
145
copy 2009 BrooksCole - Cengage
Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
copy 2009 BrooksCole - Cengage
bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
145
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Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its
ions according to
AgBr Ag+ + Br- Ksp = 5 x 10-13
bull Also Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
Solubility and Complex Ions Example
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
146
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bull Therefore the equilibrium expressed by both reactions is
AgBr Ag+ + Br- Ksp = 5 x 10-13
Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107
AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6
Keq = Ksp x Kf
bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water
bull In other words NH3 makes AgBr more soluble
Solubility and Complex Ions Example
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice
Separating Metal Ions Cu2+ Ag+ Pb2+
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separating Salts by Differences in Ksp
Separating Salts by Differences in Ksp (2)
Separations by Difference in Ksp
Qualitative Analysis
Slide 138
Reaction of NH3 with Cu2+(aq)
Solubility and Complex Ions
Solubility and Complex Ions (2)
Solubility and Complex Ions (3)
Slide 143
Solubility and Complex Ions (4)
Solubility and Complex Ions Example
Solubility and Complex Ions Example (2)
AP Exam Practice
147
copy 2009 BrooksCole - Cengage
AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1
Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1
More About Chemical Equilibria Acid-Base amp Precipitation React
Objectives ndash Chapter 15
Stomach Acidity amp Acid-Base Reactions
Acid-Base Reactions
The Common Ion Effect Section 151
Slide 6
Slide 7
Slide 8
pH of NH3NH4+ Mixture
pH of NH3NH4+ Mixture (2)
Common Ion Effect
Common Ion Effect Practice
Common Ion Effect Practice (2)
Buffered Solutions Section 152
Buffer Solutions
Buffer Solutions (2)
Buffer Solutions (3)
Buffer Solutions (4)
Buffer Solutions (5)
Buffer Solutions (6)
Buffer Solutions (7)
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation (2)
Buffer Calculation Example (AcidConj Base)
Buffer Calculation Example
Buffer Calculation using the HendersonndashHasselbalch Equation
Buffer Calculation using the HendersonndashHasselbalch Equation (2)
Buffer Practice 1
Buffer Practice 2
Adding an Acid to a Buffer (How a Buffer Maintains pH)
Adding an Acid to a Buffer
Adding an Acid to a Buffer
Adding an Acid to a Buffer (2)
Adding an Acid to a Buffer (3)
Adding an Acid to a Buffer (4)
Adding an Acid to a Buffer (5)
Adding a Base to a Buffer
Adding an Base to a Buffer
Adding a Base to a Buffer (2)
Adding a Base to a Buffer (3)
Adding a Base to a Buffer (4)
Adding a Base to a Buffer (5)
Preparing a Buffer
Preparing a Buffer (2)
Preparing a Buffer (3)
Preparing a Buffer (4)
Commercial Buffers
Preparing a Buffer (5)
Buffer Example
Buffer Example (2)
Preparing a buffer Practice
Buffering Capacity Chapter 153
Buffer Capacity
Large Buffer System
Calculate pH via Henderson-Hasselbalch Equations
Small Buffer System
Calculating pH for Small Buffer System
Calculate pH via Henderson-Hasselbalch Equations (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15
Mixing Acids and Bases (Stoichiometry amp Equilibrium)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (3)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (4)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (5)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (6)
Mixing Acids and Bases (Stoichiometry amp Equilibrium) (7)
Slide 67
Titrations and pH Curves
Titrations and pH Curves (2)
Titrations and pH Curves (3)
Slide 71
Titrations and pH Curves (4)
Titrations and pH Curves (5)
Titrations and pH Curves (6)
Slide 75
Titrations and pH Curves (7)
Titrations and pH Curves (8)
Titrations and pH Curves (9)
Titrations and pH Curves (10)
Titrations and pH Curves (11)
Slide 81
Slide 82
Titrations and pH Curves (12)
Titrations and pH Curves (13)
Titrations and pH Curves (14)
Acetic acid titrated with NaOH
Strong acid titrated with a strong base
Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)
Slide 89
Weak base (NH3) titrated with a strong acid (HCl)
Acid-Base Indicators Chapter 155
Acid-Base Indicators
Indicators for Acid-Base Titrations
Titration of a Base with an Acid
Solubility and Complex Ion Equilibria Chapter 16
Objectives ndash Chapter 16
Types of Chemical Reactions
Analysis of Silver Group
Analysis of Silver Group (2)
Analysis of Silver Group (3)
Analysis of Silver Group (4)
Analysis of Silver Group (5)
Solubility Products
Slide 104
Lead(II) Chloride
Solubility of Lead(II) Iodide
Solubility of Lead(II) Iodide (2)
Solubility of Lead(II) Iodide (3)
SolubilityKsp Practice 1
SolubilityKsp Practice 2
The Common Ion Effect
Common Ion Effect
Barium Sulfate Ksp = 11 x 10-10
The Common Ion Effect (2)
The Common Ion Effect (3)
The Common Ion Effect (4)
The Common Ion Effect (5)
The Common Ion Effect (6)
Common Ion Effect Practice (3)
Factors Affecting Solubility - pH
Solubility in more Acidic Solution example solid Mg(OH)2 in wa
Solubility in more Acidic Solution example solid Mg(OH)2 in p
Acidic solution example Ag3PO4
Acidic solution example AgCl
Solubility and Acidity
Precipitation and Qualitative Analysis Section 162
Precipitation and Qualitative Analysis
Precipitation and Qualitative Analysis (2)
Precipitation and Qualitative Analysis (3)
Precipitation and Qualitative Analysisndash Practice