John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.

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





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© 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


Stomach Acidity &Stomach Acidity &Acid-Base ReactionsAcid-Base Reactions

PLAY MOVIE

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5


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


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


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)?


pH of Aqueous NHpH of Aqueous NH33

8


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)?


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


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)?


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


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)?


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


11


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)?


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


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


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

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13


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


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-- !!!! !!!!


15


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++ ! !


16


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?


KKaa = 1.8 x 10 = 1.8 x 10-5-5


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


[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?


KKaa = 1.8 x 10 = 1.8 x 10-5-5


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


[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?


KKaa = 1.8 x 10 = 1.8 x 10-5-5


[H[H33OO++] = 2.1 x 10] = 2.1 x 10-5-5 and and pH = 4.68pH = 4.68

19


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


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


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


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



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



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


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



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)



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




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.



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




HOAc + HHOAc + H22O O ee HH33OO++ + OAc + OAc--



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


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


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


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


30


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


31


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


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?


PLAY MOVIE

33


TitrationsTitrationsTitrationsTitrations

pHpHpHpH

Titrant volume, mLTitrant volume, mLTitrant volume, mLTitrant volume, mL

34


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



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



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


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


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


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



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




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



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



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


44



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


125 mL125 mL

[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M

45



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



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


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


48




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


49



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


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


See Figure 18.4See Figure 18.4

Strong acid titrated with a strong baseStrong acid titrated with a strong base

52


Weak diprotic Weak diprotic acid (Hacid (H22CC22OO44) ) titrated with a titrated with a strong base strong base

(NaOH)(NaOH)


53


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



Weak base (NHWeak base (NH33) ) titrated with a strong titrated with a strong

acid (HCl)acid (HCl)

55


Acid-Base Acid-Base IndicatorsIndicatorsSee Figure See Figure

18.818.8

Acid-Base Acid-Base IndicatorsIndicatorsSee Figure See Figure

18.818.8

56


Indicators for Acid-Base Indicators for Acid-Base TitrationsTitrations

Indicators for Acid-Base Indicators for Acid-Base TitrationsTitrations

57


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


PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts


PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts


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Lead(II) iodide

59


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


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

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61


Analysis Analysis of Silver of Silver

GroupGroup


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



GroupGroup


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



GroupGroup


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



GroupGroup


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


Some Values of KSome Values of Kspsp

Table 18.2 and Appendix JTable 18.2 and Appendix J

66


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

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67


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


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


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



= 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


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






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






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




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


The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes

the equilibrium to shift back to reactant.

74


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

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75


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


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


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..


BaSOBaSO44(s) (s) ee BaBa2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)

78


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




79


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.





80


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!





81


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

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82


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



[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


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?


85


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


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


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


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


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


Separations by Difference in Separations by Difference in KKspsp

See Figure 18.18

90


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


Reaction of NHReaction of NH33 with Cu with Cu2+2+

(aq)(aq)

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92


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--

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93


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

John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.

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