John C. Kotz • State University of New York, College at Oneonta
John C. KotzPaul M. TreichelJohn Townsend
http://academic.cengage.com/kotz
Chapter 18Principles of Reactivity:
Other Aspects of Aqueous Equilibria
Important – Read Before Using Slides in Class
Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc.
If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. Follow these steps:
1. Go to the disc drive directory containing the PowerLecture disc, and then to the “Media” folder, and then to the
“PowerPoint_Lectures” folder. 2. In the “PowerPoint_Lectures” folder, copy the entire chapter
folder to your computer. Chapter folders are named “chapter1”, “chapter2”, etc. Each chapter folder contains the PowerPoint Lecture file as well as the animation and video files.
For assistance with installing the fonts or copying the animations and video files, please visit our Technical Support at http://academic.cengage.com/support or call (800) 423-0563. Thank you.
3
© 2009 Brooks/Cole - Cengage
More About More About Chemical EquilibriaChemical Equilibria
Acid-Base & Precipitation Acid-Base & Precipitation
ReactionsReactions Chapter 18Chapter 18
PLAY MOVIEPLAY MOVIE
4
© 2009 Brooks/Cole - Cengage
Stomach Acidity &Stomach Acidity &Acid-Base ReactionsAcid-Base Reactions
PLAY MOVIE
PLAY MOVIE
5
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
• Strong acid + strong baseStrong acid + strong base
HCl + NaOH HCl + NaOH ff• Strong acid + weak baseStrong acid + weak base
HCl + NHHCl + NH33 ff• Weak acid + strong baseWeak acid + strong base
HOAc + NaOH HOAc + NaOH ff• Weak acid + weak baseWeak acid + weak base
HOAc + NHHOAc + NH33 ff
What is relative pH What is relative pH
beforebefore, , duringduring, & , &
afterafter reaction? reaction?
Need to study:Need to study:
a) Common ion a) Common ion
effect and bufferseffect and buffers
b) Titrationsb) Titrations
6
© 2009 Brooks/Cole - Cengage
QUESTION: What is the effect on the pH of adding QUESTION: What is the effect on the pH of adding NHNH44Cl to 0.25 M NHCl to 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
Here we are adding NHHere we are adding NH44++, an ion , an ion COMMONCOMMON to the to the
equilibrium. equilibrium.
Le Chatelier predicts that the equilibrium will shift to the Le Chatelier predicts that the equilibrium will shift to the left (1), right (2), no change (3).left (1), right (2), no change (3).
The pH will go The pH will go up (1), down (2), no change (3).up (1), down (2), no change (3).
NHNH44++ is an acid! is an acid!
The Common Ion The Common Ion EffectEffect
Section 18.1Section 18.1
7
© 2009 Brooks/Cole - Cengage
Let us first calculate the pH of a 0.25 M NHLet us first calculate the pH of a 0.25 M NH33
solution.solution.
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.250.25 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.25 - x0.25 - x x x xx
QUESTION: What is the effect on the pH of adding NHQUESTION: What is the effect on the pH of adding NH44Cl to Cl to
0.25 M NH0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
pH of Aqueous NHpH of Aqueous NH33
8
© 2009 Brooks/Cole - Cengage
QUESTION: What is the effect on the pH of adding NHQUESTION: What is the effect on the pH of adding NH44Cl Cl
to 0.25 M NHto 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
Kb= 1.8 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
0.25 - x
Kb= 1.8 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x2
0.25 - x
pH of Aqueous NHpH of Aqueous NH33
Assuming x is << 0.25, we haveAssuming x is << 0.25, we have
[OH[OH--] = x = [K] = x = [Kbb(0.25)](0.25)]1/21/2 = 0.0021 M = 0.0021 M
This gives pOH = 2.67This gives pOH = 2.67
and so pH = 14.00 - 2.67and so pH = 14.00 - 2.67
= = 11.3311.33 for 0.25 M NH for 0.25 M NH33
9
© 2009 Brooks/Cole - Cengage
Problem:Problem: What is the pH of a solution with 0.10 What is the pH of a solution with 0.10 M NHM NH44Cl and 0.25 M NHCl and 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
We expect that the pH will decline on adding We expect that the pH will decline on adding NHNH44Cl. Let’s test that!Cl. Let’s test that!
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
10
© 2009 Brooks/Cole - Cengage
Problem:Problem: What is the pH of a solution with 0.10 What is the pH of a solution with 0.10 M NHM NH44Cl and 0.25 M NHCl and 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
We expect that the pH will decline on adding We expect that the pH will decline on adding NHNH44Cl. Let’s test that!Cl. Let’s test that!
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.250.25 0.100.10 00
changechange -x-x +x+x +x+x
equilibequilib 0.25 - x0.25 - x 0.10 + x 0.10 + x xx
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
11
© 2009 Brooks/Cole - Cengage
Problem:Problem: What is the pH of a solution with 0.10 M NH What is the pH of a solution with 0.10 M NH44Cl Cl
and 0.25 M NHand 0.25 M NH33(aq)?(aq)?
NHNH33(aq) + H(aq) + H22O O ee NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
Kb= 1.8 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(0.10 + x)
0.25 - x
Kb= 1.8 x 10-5 = [NH4
+ ][OH- ]
[NH3 ] =
x(0.10 + x)
0.25 - x
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
Assuming x is very small,Assuming x is very small,
[OH[OH--] = x = (0.25 / 0.10)(K] = x = (0.25 / 0.10)(Kbb) = 4.5 x 10) = 4.5 x 10-5-5 M M
This gives pOH = 4.35 and This gives pOH = 4.35 and pH = 9.65pH = 9.65pH drops from 11.33 to 9.65 pH drops from 11.33 to 9.65
on adding a common ionon adding a common ion
12
© 2009 Brooks/Cole - Cengage
Buffer SolutionsBuffer SolutionsSection 18.2Section 18.2
HCl is added to HCl is added to pure water.pure water.
HCl is added to a HCl is added to a solution of a weak solution of a weak acid Hacid H22POPO44
-- and its and its conjugate base conjugate base HPOHPO44
2-2-..
PLAY MOVIE
PLAY MOVIE
13
© 2009 Brooks/Cole - Cengage
A buffer solution is a special case of the A buffer solution is a special case of the common ion effect. common ion effect.
The function of a buffer is to resist changes The function of a buffer is to resist changes in the pH of a solution.in the pH of a solution.
Buffer CompositionBuffer Composition
Weak AcidWeak Acid ++ Conj. BaseConj. Base
HOAcHOAc ++ OAcOAc--
HH22POPO44-- ++ HPOHPO44
2-2-
NHNH44++
++ NHNH33
Buffer SolutionsBuffer Solutions
14
© 2009 Brooks/Cole - Cengage
Consider HOAc/OAcConsider HOAc/OAc-- to see how buffers work to see how buffers work
ACID USES UP ADDED OHACID USES UP ADDED OH--
We know thatWe know that
OAcOAc-- + H + H22O O ee HOAc + OHHOAc + OH--
has Khas Kbb = 5.6 x 10 = 5.6 x 10-10-10
Therefore, the Therefore, the reverse reactionreverse reaction of the WEAK of the WEAK ACID with added OHACID with added OH--
has Khas Kreversereverse = 1/ K = 1/ Kbb = = 1.8 x 101.8 x 1099
KKreversereverse is VERY LARGE, so HOAc is VERY LARGE, so HOAc completely snarfs up OHcompletely snarfs up OH-- !!!! !!!!
Buffer SolutionsBuffer Solutions
15
© 2009 Brooks/Cole - Cengage
Consider HOAc/OAcConsider HOAc/OAc-- to see how buffers to see how buffers workwork
CONJ. BASE USES UP ADDED HCONJ. BASE USES UP ADDED H++
HOAc + HHOAc + H22O O ee OAcOAc-- + H + H33OO++
has Khas Kaa = 1.8 x 10 = 1.8 x 10-5-5
Therefore, the Therefore, the reverse reactionreverse reaction of the WEAK of the WEAK BASE with added HBASE with added H++
has Khas Kreversereverse = 1/ K = 1/ Kaa = = 5.6 x 105.6 x 1044
KKreversereverse is VERY LARGE, so OAc is VERY LARGE, so OAc-- completely snarfs up Hcompletely snarfs up H++ ! !
Buffer SolutionsBuffer Solutions
16
© 2009 Brooks/Cole - Cengage
Problem:Problem: What is the pH of a buffer that has What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc[HOAc] = 0.700 M and [OAc--] = 0.600 M?] = 0.600 M?
HOAc + HHOAc + H22O O ee OAcOAc-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-5-5
Buffer SolutionsBuffer Solutions
0.700 M HOAc has pH = 2.45
The pH of the buffer will have
1. pH < 2.45
2. pH > 2.45
3. pH = 2.45
0.700 M HOAc has pH = 2.45
The pH of the buffer will have
1. pH < 2.45
2. pH > 2.45
3. pH = 2.45
17
© 2009 Brooks/Cole - Cengage
[HOAc][HOAc] [OAc[OAc--]] [H[H33OO++]]
initialinitial
changechange
equilibequilib
Problem:Problem: What is the pH of a buffer that has What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc[HOAc] = 0.700 M and [OAc--] = 0.600 M?] = 0.600 M?
HOAc + HHOAc + H22O O ee OAcOAc-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-5-5
Buffer SolutionsBuffer Solutions
0.7000.700 0.6000.600 00-x-x +x+x +x+x
0.700 - x0.700 - x 0.600 + x0.600 + x xx
18
© 2009 Brooks/Cole - Cengage
[HOAc][HOAc] [OAc[OAc--]] [H[H33OO++]]
equilibequilib 0.700 - x0.700 - x 0.600 + x0.600 + x xx
Assuming that x << 0.700 and 0.600, we haveAssuming that x << 0.700 and 0.600, we have
Ka= 1.8 x 10-5 =
[H3O+ ](0.600)
0.700
Ka= 1.8 x 10-5 =
[H3O+ ](0.600)
0.700
Problem:Problem: What is the pH of a buffer that has [HOAc] = 0.700 M What is the pH of a buffer that has [HOAc] = 0.700 M and [OAcand [OAc--] = 0.600 M?] = 0.600 M?
HOAc + HHOAc + H22O O ee OAcOAc-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-5-5
Buffer SolutionsBuffer Solutions
[H[H33OO++] = 2.1 x 10] = 2.1 x 10-5-5 and and pH = 4.68pH = 4.68
19
© 2009 Brooks/Cole - Cengage
Notice that the expression for calculating the Notice that the expression for calculating the HH++ conc. of the buffer is conc. of the buffer is
[H3O+ ] = Orig. conc. of HOAc
Orig. conc. of OAc- x Ka
[H3O+ ] = Orig. conc. of HOAc
Orig. conc. of OAc- x Ka
Buffer SolutionsBuffer Solutions
Notice that the HNotice that the H33OO++ or OH or OH-- concs. depend on (1) K and (2) concs. depend on (1) K and (2)
the ratio of acid and base the ratio of acid and base concs.concs.
[H3O+ ] =
[Acid]
[Conj. base] x Ka
[H3O+ ] =
[Acid]
[Conj. base] x Ka
[OH- ] =
[Base]
[Conj. acid] x Kb
[OH- ] =
[Base]
[Conj. acid] x Kb
20
© 2009 Brooks/Cole - Cengage
Henderson-Hasselbalch EquationHenderson-Hasselbalch Equation
Take the Take the negative lognegative log of both sides of this of both sides of this equationequation
[H3O+ ] =
[Acid]
[Conj. base] x Ka
[H3O+ ] =
[Acid]
[Conj. base] x Ka
pH = pKa + log
[Conj. base]
[Acid] pH = pKa + log
[Conj. base]
[Acid]
pH = pKa - log[ ]Acid
[ . ]Conj base
The pH is determined largely by the The pH is determined largely by the pKpKaa of the acid of the acid and and
then then adjusted by the ratio of acid and conjugate base.adjusted by the ratio of acid and conjugate base.
21
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Problem:Problem: What is the pH when 1.00 mL of 1.00 M HCl What is the pH when 1.00 mL of 1.00 M HCl is added tois added to
a)a)1.00 L of pure water (before HCl, pH = 7.00)1.00 L of pure water (before HCl, pH = 7.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M 1.00 L of buffer that has [HOAc] = 0.700 M and [OAcand [OAc--] = 0.600 M ] = 0.600 M (pH = 4.68)(pH = 4.68)
Solution to Part (a)Solution to Part (a)
Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of waterwater
CC11•V•V11 = C = C22 • V • V22
CC22 = 1.00 x 10 = 1.00 x 10-3-3 M = [H M = [H33OO++]]
pH = 3.00pH = 3.00
22
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Solution to Part (b)Solution to Part (b)
Step 1 — do the stoichiometryStep 1 — do the stoichiometry
HH33OO++ (from HCl) + OAc (from HCl) + OAc-- (from buffer) (from buffer) ff HOAc (from buffer)HOAc (from buffer)
The reaction occurs completely because K is The reaction occurs completely because K is very large.very large.
What is the pH when 1.00 mL of 1.00 M HCl is added to What is the pH when 1.00 mL of 1.00 M HCl is added to
a)a) 1.00 L of pure water (after HCl, pH = 3.00)1.00 L of pure water (after HCl, pH = 3.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc[OAc--] = 0.600 M ] = 0.600 M (pH before = 4.68)(pH before = 4.68)
23
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Solution to Part (b): Step 1—StoichiometrySolution to Part (b): Step 1—Stoichiometry
[H[H33OO++] +] + [OAc[OAc--] ] [HOAc][HOAc]
Before rxnBefore rxn
ChangeChange
After rxnAfter rxn
What is the pH when 1.00 mL of 1.00 M HCl is added to What is the pH when 1.00 mL of 1.00 M HCl is added to
a)a) 1.00 L of pure water (pH = 3.00)1.00 L of pure water (pH = 3.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc[OAc--] = 0.600 M ] = 0.600 M (pH = 4.68)(pH = 4.68)
0.00100 mol0.00100 mol0.600 mol0.600 mol 0.700 mol0.700 mol
-0.00100-0.00100 -0.00100-0.00100 +0.00100+0.00100
00 0.599 mol0.599 mol 0.701 mol0.701 mol
24
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Solution to Part (b): Step 2—EquilibriumSolution to Part (b): Step 2—Equilibrium
HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--
[HOAc] [HOAc] [H[H33OO++] [OAc] [OAc--]]
Before rxn (M)Before rxn (M)
Change (M)Change (M)
After rxn (M)After rxn (M)
What is the pH when 1.00 mL of 1.00 M HCl is added to What is the pH when 1.00 mL of 1.00 M HCl is added to
a)a) 1.00 L of pure water (pH = 3.00)1.00 L of pure water (pH = 3.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc[OAc--] = 0.600 M ] = 0.600 M (pH = 4.68)(pH = 4.68)
0.701 mol/L0.701 mol/L00 0.599 mol/L0.599 mol/L-x-x +x+x +x+x
0.599 + x0.599 + xxx0.701-x0.701-x
25
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Solution to Part (b): Step 2—EquilibriumSolution to Part (b): Step 2—Equilibrium
HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--
[HOAc] [HOAc] [H[H33OO++] ] [OAc[OAc--]]
After rxnAfter rxn 0.701-x0.701-x 0.599+x0.599+x xx
Because [HBecause [H33OO++] = 2.1 x 10] = 2.1 x 10-5-5 M BEFORE adding M BEFORE adding HCl, we again HCl, we again neglect xneglect x relative to 0.701 and relative to 0.701 and 0.599. 0.599.
What is the pH when 1.00 mL of 1.00 M HCl is added to What is the pH when 1.00 mL of 1.00 M HCl is added to
a)a) 1.00 L of pure water (pH = 3.00)1.00 L of pure water (pH = 3.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M [OAc-] = 0.600 M (pH = 4.68)(pH = 4.68)
26
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Adding an Acid to a BufferBuffer
Solution to Part (b): Step 2—EquilibriumSolution to Part (b): Step 2—Equilibrium
HOAc + HHOAc + H22O O ee HH33OO++ + OAc + OAc--
What is the pH when 1.00 mL of 1.00 M HCl is added to What is the pH when 1.00 mL of 1.00 M HCl is added to
a)a) 1.00 L of pure water (pH = 3.00)1.00 L of pure water (pH = 3.00)
b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M [OAc-] = 0.600 M (pH = 4.68)(pH = 4.68)
[H3O+ ] = [HOAc]
[OAc- ] x Ka=
0.701
0.599 x (1.8 x 10-5 )
[H3O+ ] = [HOAc]
[OAc- ] x Ka=
0.701
0.599 x (1.8 x 10-5 )
[H[H33OO++] = 2.1 x 10] = 2.1 x 10-5-5 M M ff pH = 4.68pH = 4.68The pH has not changed The pH has not changed
on adding HCl to the buffer!on adding HCl to the buffer!
27
© 2009 Brooks/Cole - Cengage
Preparing a BufferPreparing a BufferYou want to buffer a solution at pH = 4.30. You want to buffer a solution at pH = 4.30.
This means [HThis means [H33OO++] = 10] = 10-pH-pH = 5.0 x 10 = 5.0 x 10-5-5 M M
It is best to choose an acid such that [HIt is best to choose an acid such that [H33OO++] ]
is about equal to Kis about equal to Kaa (or pH ≈ pK (or pH ≈ pKaa).).
——then you get the exact [Hthen you get the exact [H33OO++] by adjusting ] by adjusting
the ratio of acid to conjugate base.the ratio of acid to conjugate base.
[H3O+ ] =
[Acid]
[Conj. base] x Ka
[H3O+ ] =
[Acid]
[Conj. base] x Ka
28
© 2009 Brooks/Cole - Cengage
You want to buffer a solution at pH = 4.30 or You want to buffer a solution at pH = 4.30 or
[H[H33OO++] = 5.0 x 10] = 5.0 x 10-5-5 M M
POSSIBLE ACIDSPOSSIBLE ACIDS KKaa
HSOHSO44- - / SO/ SO44
2-2- 1.2 x 101.2 x 10-2-2
HOAc / OAcHOAc / OAc-- 1.8 x 101.8 x 10-5-5
HCN / CNHCN / CN-- 4.0 x 104.0 x 10-10-10
Best choice is acetic acid / acetate.Best choice is acetic acid / acetate.
Preparing a BufferPreparing a Buffer
29
© 2009 Brooks/Cole - Cengage
You want to buffer a solution at pH = 4.30 or You want to buffer a solution at pH = 4.30 or
[H[H33OO++] = 5.0 x 10] = 5.0 x 10-5-5 M M
[H3O+ ] = 5.0 x 10-5 = [HOAc]
[OAc- ] (1.8 x 10-5 )
[H3O+ ] = 5.0 x 10-5 = [HOAc]
[OAc- ] (1.8 x 10-5 )
Therefore, if you use 0.100 mol of NaOAc Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = and 0.278 mol of HOAc, you will have pH = 4.30.4.30.
Solve for [HOAc]/[OAcSolve for [HOAc]/[OAc--] ratio] ratio =
2.78
1
Preparing a BufferPreparing a Buffer
30
© 2009 Brooks/Cole - Cengage
A final point —A final point —
CONCENTRATION of the acid and conjugate CONCENTRATION of the acid and conjugate base are not important.base are not important.
It is the It is the RATIO OF THE NUMBER OF MOLESRATIO OF THE NUMBER OF MOLES of each.of each.
Result: Result: diluting a buffer diluting a buffer solution does not change its solution does not change its pHpH
Preparing a BufferPreparing a Buffer
31
© 2009 Brooks/Cole - Cengage
Commercial BuffersCommercial Buffers
• The solid acid and conjugate base in the packet are mixed with water to give the specified pH.
• Note that the quantity of water does not affect the pH of the buffer.
32
© 2009 Brooks/Cole - Cengage
Buffer prepared fromBuffer prepared from
8.4 g NaHCO8.4 g NaHCO33
weak acidweak acid
16.0 g Na16.0 g Na22COCO33
conjugate baseconjugate base
HCOHCO33- - + H+ H22OO
ee H H33OO++ + CO + CO332-2-
What is the pH?What is the pH?
Preparing a BufferPreparing a Buffer
PLAY MOVIE
33
© 2009 Brooks/Cole - Cengage
TitrationsTitrationsTitrationsTitrations
pHpHpHpH
Titrant volume, mLTitrant volume, mLTitrant volume, mLTitrant volume, mL
34
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationsAcid-Base Titrations
Adding NaOH from the buret to acetic acid in the flask, a weak acid. Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.In the beginning the pH increases very slowly.
PLAY MOVIE
35
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationsAcid-Base Titrations
Additional NaOH is added. pH rises as equivalence point is Additional NaOH is added. pH rises as equivalence point is approached.approached.
PLAY MOVIE
36
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationsAcid-Base Titrations
Additional NaOH is added. pH increases and then levels off as Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.NaOH is added beyond the equivalence point.
PLAY MOVIE
37
© 2009 Brooks/Cole - Cengage
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
pH at pH at equivalence equivalence point?point?
pH at pH at equivalence equivalence point?point?
pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acidweak acid
pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acidweak acid
Benzoic acid Benzoic acid + NaOH+ NaOHBenzoic acid Benzoic acid + NaOH+ NaOH
pH at pH at half-way half-way point?point?
pH at pH at half-way half-way point?point?
38
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationAcid-Base TitrationSection 18.3Section 18.3
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution? solution?
HBz + NaOH HBz + NaOH f Na Na++ + Bz + Bz-- + H + H22OO
CC66HH55COCO22H = HBzH = HBz Benzoate ion = Bz-
39
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationAcid-Base TitrationSection 18.3Section 18.3
The pH of the final solution will beThe pH of the final solution will be1.1. Less than 7Less than 72.2. Equal to 7Equal to 73.3. Greater than 7Greater than 7
The pH of the final solution will beThe pH of the final solution will be1.1. Less than 7Less than 72.2. Equal to 7Equal to 73.3. Greater than 7Greater than 7
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the the equivalence point. What is the pH of the final solution? final solution?
HBz + NaOH HBz + NaOH f Na Na++ + Bz + Bz-- + H + H22OO
40
© 2009 Brooks/Cole - Cengage
Acid-Base TitrationsAcid-Base Titrations
The product of the titration of benzoic acid is the The product of the titration of benzoic acid is the benzoate ion, Bzbenzoate ion, Bz-- . .
BzBz-- is the conjugate base of a weak acid. is the conjugate base of a weak acid.
Therefore, final solution is basic.Therefore, final solution is basic.
BzBz- - + H+ H22O O ee HBz + OH HBz + OH--
KKbb = 1.6 x 10 = 1.6 x 10-10-10
++ee
41
© 2009 Brooks/Cole - Cengage
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
pH at pH at equivalence equivalence point is point is basicbasic
pH at pH at equivalence equivalence point is point is basicbasic
Benzoic acid Benzoic acid + NaOH+ NaOHBenzoic acid Benzoic acid + NaOH+ NaOH
42
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
Strategy —Strategy — find the conc. of the conjugate find the conc. of the conjugate base Bzbase Bz-- in the solution AFTER the in the solution AFTER the titration, then calculate pH.titration, then calculate pH.
This is a two-step problemThis is a two-step problem
1.1. stoichiometrystoichiometry of acid-base reaction of acid-base reaction
2.2. equilibrium calculationequilibrium calculation
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
43
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d1. Calc. moles of NaOH req’d
(0.100 L HBz)(0.025 M) = 0.0025 mol HBz(0.100 L HBz)(0.025 M) = 0.0025 mol HBz
This requires This requires 0.0025 mol NaOH0.0025 mol NaOH
2.2. Calc. volume of NaOH req’dCalc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L0.0025 mol (1 L / 0.100 mol) = 0.025 L25 mL of NaOH req’d25 mL of NaOH req’d
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
44
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION25 mL of NaOH req’d 25 mL of NaOH req’d 3. Moles of Bz3. Moles of Bz-- produced = moles HBz = produced = moles HBz =
0.0025 mol0.0025 mol4. Calc. conc. of Bz4. Calc. conc. of Bz--
There are 0.0025 mol of BzThere are 0.0025 mol of Bz-- in a in a TOTAL TOTAL SOLUTION VOLUMESOLUTION VOLUME of of
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
125 mL125 mL
[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M
45
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
Equivalence PointEquivalence PointMost important species in solution is benzoate Most important species in solution is benzoate
ion, Bzion, Bz--, the weak conjugate base of benzoic , the weak conjugate base of benzoic acid, HBz. acid, HBz.
BzBz- - + H+ H22O O ee HBz + OH HBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]initialinitial 0.020 0.020 00 00changechange- x- x +x+x +x +x equilibequilib 0.020 - x0.020 - x xx xx
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point?What is the pH at equivalence point?
46
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
Equivalence PointEquivalence Point
Most important species in solution is benzoate ion, BzMost important species in solution is benzoate ion, Bz--, , the weak conjugate base of benzoic acid, HBz.the weak conjugate base of benzoic acid, HBz.
BzBz- - + H+ H22O O ee HBz + OH HBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point?What is the pH at equivalence point?
x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6
pOH = 5.75 and pOH = 5.75 and pH = 8.25pH = 8.25
Kb = 1.6 x 10-10 =
x2
0.020 - x Kb = 1.6 x 10-10 =
x2
0.020 - x
47
© 2009 Brooks/Cole - Cengage
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
pH at half-way point?pH at half-way point?1.1. < 7< 72.2. = 7= 73.3. > 7> 7
pH at half-way point?pH at half-way point?1.1. < 7< 72.2. = 7= 73.3. > 7> 7
Equivalence point Equivalence point pH = 8.25pH = 8.25
Equivalence point Equivalence point pH = 8.25pH = 8.25
48
© 2009 Brooks/Cole - Cengage
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
pH at half-pH at half-way pointway point
pH at half-pH at half-way pointway point Equivalence point Equivalence point
pH = 8.25pH = 8.25
Equivalence point Equivalence point pH = 8.25pH = 8.25
49
© 2009 Brooks/Cole - Cengage
Acid-Base ReactionsAcid-Base Reactions
HBzHBz + H+ H22O O ee H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5HBzHBz + H+ H22O O ee H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5
You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH.benzoic acid with 0.100 M NaOH.
What is the pH at the half-way What is the pH at the half-way point?point?
[H3O+ ] = [HBz]
[Bz- ] x Ka
[H3O+ ] = [HBz]
[Bz- ] x Ka
At the half-way point, At the half-way point, [HBz] = [Bz[HBz] = [Bz--]]Therefore, [HTherefore, [H33OO++] = K] = Kaa = 6.3 x 10 = 6.3 x 10-5-5
pH = 4.20 = pKpH = 4.20 = pKaa of the acid of the acid
Both HBz and BzBoth HBz and Bz-- are present.are present.
This is a BUFFER!This is a BUFFER!
Both HBz and BzBoth HBz and Bz-- are present.are present.
This is a BUFFER!This is a BUFFER!
50
© 2009 Brooks/Cole - Cengage
Acetic acid titrated with NaOHAcetic acid titrated with NaOH
See Fig 18.5: Weak acid See Fig 18.5: Weak acid titrated with a strong basetitrated with a strong base
51
© 2009 Brooks/Cole - Cengage
See Figure 18.4See Figure 18.4
Strong acid titrated with a strong baseStrong acid titrated with a strong base
52
© 2009 Brooks/Cole - Cengage
Weak diprotic Weak diprotic acid (Hacid (H22CC22OO44) ) titrated with a titrated with a strong base strong base
(NaOH)(NaOH)
See Figure 18.6See Figure 18.6
53
© 2009 Brooks/Cole - Cengage
Titration of aTitration of a1. Strong acid with strong base?1. Strong acid with strong base?2. Weak acid with strong base?2. Weak acid with strong base?3. Strong base with weak acid?3. Strong base with weak acid?4. Weak base with strong acid?4. Weak base with strong acid?5. Weak base with weak acid5. Weak base with weak acid6. Weak acid with weak base?6. Weak acid with weak base?
pHpH
Volume of titrating reagent added -->Volume of titrating reagent added -->
54
© 2009 Brooks/Cole - Cengage
See Figure 18.7See Figure 18.7
Weak base (NHWeak base (NH33) ) titrated with a strong titrated with a strong
acid (HCl)acid (HCl)
55
© 2009 Brooks/Cole - Cengage
Acid-Base Acid-Base IndicatorsIndicatorsSee Figure See Figure
18.818.8
Acid-Base Acid-Base IndicatorsIndicatorsSee Figure See Figure
18.818.8
56
© 2009 Brooks/Cole - Cengage
Indicators for Acid-Base Indicators for Acid-Base TitrationsTitrations
Indicators for Acid-Base Indicators for Acid-Base TitrationsTitrations
57
© 2009 Brooks/Cole - Cengage
Natural IndicatorsNatural IndicatorsRed rose extract at different pH’s and with AlRed rose extract at different pH’s and with Al3+3+ ions ions
Add HClAdd HCl Add NHAdd NH33 Add NHAdd NH33/NH/NH44++
Add AlAdd Al3+3+
Rose extractRose extractIn CHIn CH33OHOH
58
© 2009 Brooks/Cole - Cengage
PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts
Section 18.4Section 18.4
PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts
Section 18.4Section 18.4
PLAY MOVIE
Lead(II) iodide
59
© 2009 Brooks/Cole - Cengage
Types of Chemical Types of Chemical ReactionsReactions
• EXCHANGE REACTIONS: EXCHANGE REACTIONS: AABB + C + CDD ff A ADD + C + CBB
– Acid-base: Acid-base: CHCH33COCO22H + NaOH H + NaOH ff NaCH NaCH33COCO22 + H + H22OO
– Gas forming: Gas forming: CaCOCaCO33 + 2 HCl + 2 HCl ff CaCl CaCl22 + CO + CO22(g)(g) + H + H22OO
– Precipitation: Precipitation: Pb(NOPb(NO33) ) 22 + 2 KI + 2 KI ff PbI PbI22(s)(s) + 2 KNO + 2 KNO33
• OXIDATION REDUCTIONOXIDATION REDUCTION– 4 Fe + 3 O4 Fe + 3 O22 ff 2 Fe 2 Fe22OO33
• Apply equilibrium principles to acid-base and Apply equilibrium principles to acid-base and precipitation reactions.precipitation reactions.
60
© 2009 Brooks/Cole - Cengage
Analysis of Silver Analysis of Silver GroupGroup
Analysis of Silver Analysis of Silver GroupGroup
All salts formed in All salts formed in this experiment are this experiment are said to be said to be INSOLUBLEINSOLUBLE and and form when mixing form when mixing moderately moderately concentrated concentrated solutions of the solutions of the metal ion with metal ion with chloride ions.chloride ions.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
PLAY MOVIE
61
© 2009 Brooks/Cole - Cengage
Analysis Analysis of Silver of Silver
GroupGroup
Analysis Analysis of Silver of Silver
GroupGroupAlthough all salts formed in this experiment are Although all salts formed in this experiment are
said to be insoluble, they do dissolve to some said to be insoluble, they do dissolve to some
SLIGHT extent.SLIGHT extent.
AgCl(s) AgCl(s) ee AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)
When equilibrium has been established, no more When equilibrium has been established, no more
AgCl dissolves and the solution is AgCl dissolves and the solution is SATURATEDSATURATED..
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
62
© 2009 Brooks/Cole - Cengage
Analysis Analysis of Silver of Silver
GroupGroup
Analysis Analysis of Silver of Silver
GroupGroup
AgCl(s) AgCl(s) ee Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
When solution is When solution is SATURATEDSATURATED, expt. shows that , expt. shows that [Ag[Ag++] = 1.67 x 10] = 1.67 x 10-5-5 M. M.
This is equivalent to the This is equivalent to the SOLUBILITYSOLUBILITY of AgCl.of AgCl.
What is [ClWhat is [Cl--]?]?
[Cl[Cl--] = [Ag] = [Ag++] = ] = 1.67 x 101.67 x 10-5-5 M M
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
63
© 2009 Brooks/Cole - Cengage
Analysis Analysis of Silver of Silver
GroupGroup
Analysis Analysis of Silver of Silver
GroupGroupAgCl(s) AgCl(s) ee Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
Saturated solution has Saturated solution has
[Ag[Ag++] = [Cl] = [Cl--] = 1.67 x 10] = 1.67 x 10-5-5 M M
Use this to calculate KUse this to calculate Kcc
KKcc = [Ag = [Ag++] [Cl] [Cl--]]
= (1.67 x 10= (1.67 x 10-5-5)(1.67 x 10)(1.67 x 10-5-5) )
= 2.79 x 10= 2.79 x 10-10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
64
© 2009 Brooks/Cole - Cengage
Analysis Analysis of Silver of Silver
GroupGroup
Analysis Analysis of Silver of Silver
GroupGroup
AgCl(s) AgCl(s) ee Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
KKcc = [Ag = [Ag++] [Cl] [Cl--] = 2.79 x 10] = 2.79 x 10-10-10
Because this is the product of “solubilities”, we call it Because this is the product of “solubilities”, we call it
KKspsp = solubility product = solubility product
constantconstant
• • See Table 18.2 and Appendix JSee Table 18.2 and Appendix J
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
65
© 2009 Brooks/Cole - Cengage
Some Values of KSome Values of Kspsp
Table 18.2 and Appendix JTable 18.2 and Appendix J
66
© 2009 Brooks/Cole - Cengage
Lead(II) ChlorideLead(II) ChloridePbClPbCl22(s) (s) ee Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)
KKspsp = 1.9 x 10 = 1.9 x 10-5-5 = [Pb = [Pb2+2+][Cl][Cl––]]22
PLAY MOVIE
67
© 2009 Brooks/Cole - Cengage
SolutionSolution
1. Solubility = [Pb1. Solubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M
[I[I--] = ?] = ?
[I[I--] = 2 x [Pb] = 2 x [Pb2+2+] = 2.60 x 10] = 2.60 x 10-3-3 M M
Solubility of Lead(II) Solubility of Lead(II) IodideIodide
Solubility of Lead(II) Solubility of Lead(II) IodideIodide
Consider PbIConsider PbI22 dissolving in water dissolving in water
PbIPbI22(s) (s) ee PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq)(aq)
Calculate KCalculate Kspsp
if solubility = 0.00130 Mif solubility = 0.00130 M
68
© 2009 Brooks/Cole - Cengage
SolutionSolution
2. K2. Kspsp = [Pb = [Pb2+2+] [I] [I--]]2 2
= [Pb= [Pb2+2+] {2 • [Pb] {2 • [Pb2+2+]}]}22
KKspsp = 4 [Pb = 4 [Pb2+2+]]33
Solubility of Lead(II) Solubility of Lead(II) IodideIodide
Solubility of Lead(II) Solubility of Lead(II) IodideIodide
= 4 (solubility)= 4 (solubility)33= 4 (solubility)= 4 (solubility)33
KKspsp = 4 (1.30 x 10 = 4 (1.30 x 10-3-3))33 = 8.79 x 10 = 8.79 x 10-9-9KKspsp = 4 (1.30 x 10 = 4 (1.30 x 10-3-3))33 = 8.79 x 10 = 8.79 x 10-9-9
Consider PbIConsider PbI22 dissolving in water dissolving in water
PbIPbI22(s) (s) ee PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq)(aq)
Calculate KCalculate Kspsp
if solubility = 0.00130 Mif solubility = 0.00130 M
69
© 2009 Brooks/Cole - Cengage
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
HgHg22ClCl22(s) (s) ee Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
If [HgIf [Hg222+2+] = 0.010 M, what [Cl] = 0.010 M, what [Cl--] is req’d to just ] is req’d to just
begin the precipitation of Hgbegin the precipitation of Hg22ClCl22??
That is, what is the maximum [ClThat is, what is the maximum [Cl--] that can be ] that can be
in solution with 0.010 M Hgin solution with 0.010 M Hg222+2+ without without
forming Hgforming Hg22ClCl22??
70
© 2009 Brooks/Cole - Cengage
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
HgHg22ClCl22(s) (s) ee Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
Recognize thatRecognize that
KKspsp = product of = product of maximum ion concs.maximum ion concs.
Precip. begins when product of Precip. begins when product of
ion concs. EXCEEDS the Kion concs. EXCEEDS the Kspsp..
71
© 2009 Brooks/Cole - Cengage
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
HgHg22ClCl22(s) (s) ee Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
SolutionSolution
[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 M,] = 0.010 M,
[Cl- ] =
Ksp
0.010 = 1.1 x 10-8 M
[Cl- ] =
Ksp
0.010 = 1.1 x 10-8 M
If this conc. of ClIf this conc. of Cl-- is just exceeded, Hg is just exceeded, Hg22ClCl22
begins to precipitate.begins to precipitate.
72
© 2009 Brooks/Cole - Cengage
Precipitating an Insoluble Precipitating an Insoluble SaltSalt
HgHg22ClCl22(s) (s) ee Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18
Now raise [ClNow raise [Cl--] to 1.0 M. What is the value of ] to 1.0 M. What is the value of [Hg[Hg22
2+2+] at this point?] at this point?
SolutionSolution
[Hg[Hg222+2+] = K] = Ksp sp / [Cl/ [Cl--]]22
= K= Kspsp / (1.0) / (1.0)22 = 1.1 x 10 = 1.1 x 10-18-18 M M
The concentration of HgThe concentration of Hg222+2+ has been reduced has been reduced
by 10by 101616 ! !
73
© 2009 Brooks/Cole - Cengage
The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes
the equilibrium to shift back to reactant.
74
© 2009 Brooks/Cole - Cengage
Common Ion EffectCommon Ion Effect
PbClPbCl22(s) (s) ee Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)
KKspsp = 1.9 x 10 = 1.9 x 10-5-5
PLAY MOVIE
75
© 2009 Brooks/Cole - Cengage
Barium Barium
SulfateSulfate
KKspsp = 1.1 x 10 = 1.1 x 10-10-10
(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.
76
© 2009 Brooks/Cole - Cengage
Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure in (a) pure water and (b) in 0.010 M Ba(NOwater and (b) in 0.010 M Ba(NO33))22..
KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10
BaSOBaSO44(s) (s) ee Ba Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
SolutionSolution
Solubility in pure water = [BaSolubility in pure water = [Ba2+2+] = [SO] = [SO442-2-] = x] = x
KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = x] = x22
x = (Kx = (Kspsp))1/21/2 = 1.1 x 10 = 1.1 x 10-5-5 M M
Solubility in pure water = 1.0 x 10Solubility in pure water = 1.0 x 10-5-5 mol/L mol/L
The Common Ion EffectThe Common Ion Effect
77
© 2009 Brooks/Cole - Cengage
SolutionSolution
Solubility in pure water = 1.1 x 10Solubility in pure water = 1.1 x 10-5-5 mol/L. mol/L.
Now dissolve BaSONow dissolve BaSO44 in water already in water already containing 0.010 M Bacontaining 0.010 M Ba2+2+. .
Which way will the “common ion” shift the Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSOequilibrium? ___ Will solubility of BaSO44 be be less than or greater than in pure water?___less than or greater than in pure water?___
The Common Ion EffectThe Common Ion EffectCalculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water and in (a) pure water and
(b) in 0.010 M Ba(NO(b) in 0.010 M Ba(NO33))22..
KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10
BaSOBaSO44(s) (s) ee BaBa2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
78
© 2009 Brooks/Cole - Cengage
SolutionSolution
[Ba[Ba2+2+]] [SO[SO442-2-]]
initialinitial
changechange
equilib.equilib.
The Common Ion EffectThe Common Ion Effect
+ y+ y0.0100.010 00
+ y+ y
0.010 + y0.010 + y yy
Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water and in (a) pure water and
(b) in 0.010 M Ba(NO(b) in 0.010 M Ba(NO33))22..
KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10
BaSOBaSO44(s) (s) ee BaBa2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
79
© 2009 Brooks/Cole - Cengage
SolutionSolution
KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = (0.010 + y) (y)] = (0.010 + y) (y)
Because y < 1.1 x 10Because y < 1.1 x 10-5-5 M (= x, the solubility in M (= x, the solubility in pure water), this means 0.010 + y is about pure water), this means 0.010 + y is about equal to 0.010. Therefore,equal to 0.010. Therefore,
KKspsp = 1.1 x 10 = 1.1 x 10-10-10 = (0.010)(y) = (0.010)(y)
y = 1.1 x 10y = 1.1 x 10-8-8 M = solubility in presence of M = solubility in presence of added Baadded Ba2+2+ ion. ion.
The Common Ion EffectThe Common Ion EffectCalculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water and in (a) pure water and
(b) in 0.010 M Ba(NO(b) in 0.010 M Ba(NO33))22..
KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10
BaSOBaSO44(s) (s) ee BaBa2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
80
© 2009 Brooks/Cole - Cengage
SUMMARYSUMMARY
Solubility in pure water = x = 1.1 x 10Solubility in pure water = x = 1.1 x 10 -5-5 M M
Solubility in presence of added BaSolubility in presence of added Ba2+2+ = 1.1 x 10= 1.1 x 10-8-8 M M
Le Chatelier’s Principle is followed!Le Chatelier’s Principle is followed!
The Common Ion EffectThe Common Ion EffectCalculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water and in (a) pure water and
(b) in 0.010 M Ba(NO(b) in 0.010 M Ba(NO33))22..
KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10
BaSOBaSO44(s) (s) ee BaBa2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
81
© 2009 Brooks/Cole - Cengage
Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Ksp Values
AgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
Ksp Values
AgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
PLAY MOVIE
82
© 2009 Brooks/Cole - Cengage
Separating Salts by Differences in Separating Salts by Differences in KKspsp
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add CrO. Add CrO442-2-
to precipitate red Agto precipitate red Ag22CrOCrO44 and yellow PbCrO and yellow PbCrO44. . Which precipitates first?Which precipitates first?
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14
SolutionSolution
The substance whose KThe substance whose Kspsp is first is first exceeded precipitates first. exceeded precipitates first.
The ion requiring the lesser amount of The ion requiring the lesser amount of CrOCrO44
2-2- ppts. first. ppts. first.
83
© 2009 Brooks/Cole - Cengage
Separating Salts by Differences in Separating Salts by Differences in KKspsp
[CrO[CrO442-2-] to ppt. PbCrO] to ppt. PbCrO4 4 = K= Ksp sp / [Pb/ [Pb2+2+] ]
= 1.8 x 10= 1.8 x 10-14-14 / 0.020 = 9.0 x 10 / 0.020 = 9.0 x 10-13-13 M M
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add CrO. Add CrO442-2- to precipitate to precipitate
red Agred Ag22CrOCrO44 and yellow PbCrO and yellow PbCrO44. Which precipitates first?. Which precipitates first?
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14
SolutionSolutionCalculate [CrOCalculate [CrO44
2-2-] required by each ion.] required by each ion.
[CrO[CrO442-2-] to ppt. Ag] to ppt. Ag22CrOCrO4 4 = K= Ksp sp / [Ag/ [Ag++]]22
= 9.0 x 10= 9.0 x 10-12-12 / (0.020) / (0.020)22 = 2.3 x 10 = 2.3 x 10-8-8 M M
PbCrOPbCrO44 precipitates first precipitates first
84
© 2009 Brooks/Cole - Cengage
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. PbCrO. PbCrO44 ppts. first. ppts. first.
KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12
KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?
Separating Salts by Differences in Separating Salts by Differences in KKspsp
85
© 2009 Brooks/Cole - Cengage
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add CrO. Add CrO442-2- to precipitate to precipitate
red Agred Ag22CrOCrO44 and yellow PbCrO and yellow PbCrO44. .
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate? begins to precipitate?
SolutionSolution
Separating Salts by Differences in Separating Salts by Differences in KKspsp
We know that [CrOWe know that [CrO442-2-] = 2.3 x 10] = 2.3 x 10-8-8 M to begin to M to begin to
ppt. Agppt. Ag22CrOCrO44. .
What is the PbWhat is the Pb2+2+ conc. at this point? conc. at this point?
[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M
= = 7.8 x 107.8 x 10-7-7 M M
Lead ion has dropped from 0.020 M to < 10Lead ion has dropped from 0.020 M to < 10-6-6 M M
86
© 2009 Brooks/Cole - Cengage
Separating Salts Separating Salts by Differences by Differences in Kin Kspsp
Separating Salts Separating Salts by Differences by Differences in Kin Kspsp
• Add CrOAdd CrO442-2- to solid PbCl to solid PbCl22. The less soluble salt, PbCrO. The less soluble salt, PbCrO44, ,
precipitatesprecipitates
• PbCl2(s) + CrO42- ee PbCrO4 + 2 Cl-
• Salt Ksp
PbCl2 1.7 x 10-5
PbCrO4 1.8 x 10-14
87
© 2009 Brooks/Cole - Cengage
Separating Salts by Separating Salts by Differences in KDifferences in Kspsp
• PbClPbCl22(s) + CrO(s) + CrO442-2- ee PbCrO PbCrO44 + 2 Cl + 2 Cl--
SaltSalt KKspsp
PbClPbCl22 1.7 x 101.7 x 10-5-5
PbCrOPbCrO44 1.8 x 101.8 x 10-14-14
PbClPbCl22(s) (s) ee Pb Pb2+2+ + 2 Cl + 2 Cl-- KK11 = K = Kspsp
PbPb2+2+ + CrO + CrO442-2- ee PbCrO PbCrO44 KK22 = 1/K = 1/Kspsp
KKnetnet = (K = (K11)(K)(K22) = 9.4 x 10) = 9.4 x 1088
Net reaction is product-favoredNet reaction is product-favored
88
© 2009 Brooks/Cole - Cengage
Lead Chemistry• From Chemistry
& Chemical
Reactivity, 5th
edition
• Illustrates
the
transformation
of one
insoluble
compound into
an even less
soluble
compound.
PbClPbCl22 PbIPbI22
Pb(COPb(CO33))
22
PbCrOPbCrO
44
89
© 2009 Brooks/Cole - Cengage
Separations by Difference in Separations by Difference in KKspsp
See Figure 18.18
90
© 2009 Brooks/Cole - Cengage
The combination of metal ions (Lewis acids) with The combination of metal ions (Lewis acids) with Lewis bases such as HLewis bases such as H22O and NHO and NH33 leads to leads to COMPLEX COMPLEX
IONSIONS
Solubility and Complex IonsSolubility and Complex IonsSections 18.6 & 18.7Sections 18.6 & 18.7
91
© 2009 Brooks/Cole - Cengage
Reaction of NHReaction of NH33 with Cu with Cu2+2+
(aq)(aq)
PLAY MOVIE
PLAY MOVIE
92
© 2009 Brooks/Cole - Cengage
Formation of complex ions explains why you Formation of complex ions explains why you can dissolve a ppt. by forming a complex can dissolve a ppt. by forming a complex ion. ion.
Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions
AgCl(s) + 2 NHAgCl(s) + 2 NH33 ee Ag(NH Ag(NH33))22+ + + Cl+ Cl--
PLAY MOVIEPLAY MOVIE
93
© 2009 Brooks/Cole - Cengage
Examine the solubility of AgCl in ammonia. Examine the solubility of AgCl in ammonia.
AgCl(s) AgCl(s) ee AgAg++ + Cl + Cl-- KKspsp = 1.8 x 10 = 1.8 x 10-10-10
AgAg++ + 2 NH + 2 NH33 ee Ag(NH Ag(NH33))22++ K Kformform = 1.6 x 10 = 1.6 x 1077
--------------------------------------------------------------------------
AgCl(s) + 2 NHAgCl(s) + 2 NH33 ee Ag(NHAg(NH33))22+ + + Cl+ Cl--
KKnetnet = (K = (Kspsp)(K)(Kformform) = 2.9 x 10) = 2.9 x 10-3-3
By adding excess NHBy adding excess NH33, the equilibrium shifts to the , the equilibrium shifts to the
right.right.
Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions