15 - 1 Ionic Equilibrium When a slightly soluble or insoluble salt is mixed with water, a saturated...

15 - 1

Ionic EquilibriumIonic EquilibriumWhen a slightly soluble or insoluble salt ismixed with water, a saturated solution quicklyresults and a dynamic equilibrium results.

This is a dynamic equilibrium because therate of the forward reaction (dissociation ofthe salt) equals the rate of the reversereaction (salt precipitating).

Reactants are converted to products at the same rate as products are converted to reactants.

15 - 2

For example, when solutions of AgNO3 andK2CrO4 are mixed, the insoluble salt,Ag2CrO4, is immediately formed.This is shown in the following equations:Ag2CrO4(s) → 2Ag+(aq) + CrO4

2-(aq) (1)2Ag+(aq) + CrO4

2-(aq) → Ag2CrO4(s) (2) Equation (1) represents Ag2CrO4 initially putinto water and dissolving.As Ag2CrO4 continues to dissolve, some of theions start to precipitate shown by Equation(2).

15 - 3

At this point the salt is dissolving at a faster rate than the precipitate is forming.

Eventually the rate of dissolving will equal the

rate of precipitating and the system will be in

state of dynamic equilibrium shown in thefollowing equation:Ag2CrO4(s) 2Ag+(aq) + CrO4

2-(aq)

The solubility product constant and thesolubility product equilibrium expression isgiven by:

15 - 4

Ksp(Ag2CrO4) = [Ag+]2[CrO42-] = 1.2 × 10-12

Ksp is used for a saturated solution of a slightly soluble salt at a given temperature. The smaller the Ksp, the less soluble the salt. Remember that solids are never included in the equilibrium expression.

15 - 5

The concentration units are moles per liter (M) and the concentration of the precipitate is included in the Ksp value.

It is important not to confuse the termssolubility product and solubility.

Solubility product is an equilibrium constant related to the equilibrium between a solid salt and its ions in solution. Solubility is the amount of solute

needed to form a saturated solution.

15 - 6

Solubility and Solubility Solubility and Solubility ProductProduct

The solubility of lead(II) sulfate is 0.038 g/L.Determine the solubility product for lead(II)sulfate.

s(PbSO4) = 0.038 g/L Ksp = ?

[PbSO4] = nV

[PbSO4] = 0.038 g PbSO4L ×

1 mol PbSO4303.27 g PbSO4

=1.2 × 10-4 M

15 - 7

PbSO4(s) Pb2+(aq) + SO42-(aq)

[ ]i 0 0[ ]c +1.2 × 10-4 +1.2 × 10-4

[ ]e 1.2 × 10-4 1.2 × 10-4

Ksp = [Pb2+][SO42-]

Ksp = 1.2 × 10-4 × 1.2 × 10-4 = 1.4 x 10-8

15 - 8

The Ksp(Pb(IO3)2) = 2.5 x 10-13.

(a)Determine its solubility in pure water.

Pb(IO3)2(s) Pb2+(aq) + 2IO3-(aq)

[ ]i 0 0[ ]c +x +2x[ ]e x 2x

Ksp = [Pb2+][IO3-]2

2.5 x 10-13 = x (2x)2× = 4x3

15 - 9

x = 4.0 x 10-5 M

s = 4.0 x 10-5 mol Pb(IO3)2L

(b) Determine the solubility in g/L in pure water.

s = 4.0 x 10-5 mol Pb(IO3)2L ×

557.00 g Pb(IO3)21 mol Pb(IO3)2

= 2.2 x 10-2 g Pb(IO3)2L

15 - 10

Common-Ion EffectCommon-Ion EffectThe common-ion effect makes it possible toshift the equilibrium to favor the reactants.

The solubility of a solid is lowered if the

solution contains an ion that is common

to the salt. The saturated solution will contain

more undissolved solid and a lower concentration of the other ions.

15 - 11

Using silver chloride as an example, theequilibrium existing in pure water would begiven by:

AgCl(s) Ag+(aq) + Cl-(aq)

In deionized water, the [Ag+] and [Cl-] wouldattain the maximum concentration dictated

byits Ksp.

But what if you tried using an aqueoussolution of sodium chloride?

15 - 12

The sodium chloride dissolves given by:

NaCl(aq) → Na+(aq) + Cl-(aq)

Cl- being the common ion between theinsoluble salt and the aqueous solution,would suppress the amount of Cl- comingfrom the AgCl.

The common-ion effect is Le Chatelier’sprinciple applied to ionic equilibria.

15 - 13

(a)Calculate the solubility in g/L of Mg3(PO4)2 in pure water if its Ksp = 1 x 10-24.

Mg3(PO4)2(s) 3Mg2+(aq) + 2PO43-(aq)

[ ]i 0 0[ ]c +3x +2x[ ]e 3x 2x

Ksp = [Mg2+]3[PO43-]2

1 x 10-24 = (3x)3 × (2x)2 = 108x5

x = 6 × 10-6 M

15 - 14

.

s = 6 × 10-6 mol Mg3(PO4)2L

×

262.84 g Mg3(PO4)21 mol Mg3(PO4)2

2 × 10-3 g Mg3(PO4)2s =

15 - 15

(b) Calculate the solubility in g/L of Mg3(PO4)2

in 0.010 M Mg(NO3)2.

Mg3(PO4)2(s) 3Mg2+(aq) + 2PO43-(aq)

[ ]i 0.010 0[ ]c +3x + 0.010 +2x[ ]e 3x + 0.010 2x

Ksp = [Mg2+]3[PO43-]2

1 x 10-24 = (3x + 0.010)3 × (2x)2

15 - 16

Chemists must bond together and continue tolook for legitimate shortcuts!

The expression (3x + 0.010)3 would be prettynasty to expand and even more nasty tocalculate!

We use the approximation 3x + 0.010 ≈ 0.010to greatly simplify matters.

The task becomes much more manageablenow that

1 x 10-24 = (0.010)3 × (2x)2

15 - 17

x = 5 x 10-10 M

s = 5 x 10-10 mol Mg3(PO4)2

L×

262.84 g Mg3(PO4)21 mol Mg3(PO4)2

s = 1 x 10-7 g Mg3(PO4)2/LThis drastic drop in solubility from 2 x 10-3 to1 x 10-7 g/L is exactly what is predicted fromLe Chatelier’s principle.

15 - 18

(c) Calculate the solubility in g/L of Mg3(PO4)2 in 0.020 M Na3PO4.

Mg3(PO4)2(s) 3Mg2+(aq) + 2PO43-(aq)

[ ]i 0 0.020[ ]c +3x +0.020 + 2x[ ]e 3x 0.020 + 2x

Ksp = [Mg2+]3[PO43-]2

1 x 10-24 = (3x)3 × (0.020 + 2x)2

(0.020 + 2x)2 ≈ 0.0202

15 - 19

1 x 10-24 = (3x)3 × (0.020)2

x = 5 x 10-8 M

s = 5 x 10-8 mol Mg3(PO4)2

L×

262.84 g Mg3(PO4)21 mol Mg3(PO4)2

s =1 x 10-5 g Mg3(PO4)2/L

15 - 20

Selective PrecipitationSelective PrecipitationThe difference in the solubility of two saltscontaining a common anion can be used toseparate a pair of cations.

The difference in the solubility of two saltscontaining a common cation can be used toseparate a pair of anions.

To illustrate this method, assume you have a

solution of sodium chloride and potassiumchromate.

15 - 21

Your task is to remove the chloride from the solution. What could you do?

One thing you could do is to boil the solutionin a beaker covered with a watch glass toprevent splattering.

The solid remaining on the bottom of the

beaker is likely to be a mixture of Na2CrO4, KCl, NaCl, and K2CrO4.

15 - 22

An alternative method is using selectiveprecipitation.

A salt whose cation will form a precipitatewith Cl- and CrO4

2- is AgNO3.

For a given problem, one would determinewhich anion would required the smallestconcentration of Cl- to form the precipitate.

15 - 23

Selective Precipitation Selective Precipitation ProblemProblem

A solution is prepared by using 0.10 M NaCland 0.010 M K2CrO4. AgNO3 is slowly addedto the solution. Ksp(AgCl) = 1.6 x 10-10 andKsp(Ag2CrO4) = 9.0 x 10-12.

(a)Which precipitates first, the AgCl or the Ag2CrO4?

AgCl(s) Ag+(aq) + Cl-(aq)[ ]i 0 0.10[ ]c +x +x[ ]e x 0.10+x

15 - 24

Ksp = [Ag+][Cl-]

1.6 x 10-10 = x (0.10+x) ≈ 0.10x

[Ag+] = 1.6 x 10-9 M

Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq)

[ ]i 0 0.010[ ]c +2x 0.010 +

x[ ]e 2x 0.010 +

x

×

15 - 25

Ksp = [Ag+]2[CrO42-]

9.0 x 10-12 = 4x2 (0.010 + x) ≈ 0.040x2

[Ag+] = 1.5 x 10-5 M × 2 = 3.0 x 10-5 M

The AgCl will precipitate first because it requires the smaller [Ag+] = 1.6 x 10-9 M.

(b) What is the [Ag+] when precipitation begins?

[Ag+] = 1.6 x 10-9 M

×

15 - 26

(c) What is [Cl-] when the Ag2CrO4 starts to precipitate?

Ksp = [Ag+]2[CrO42-]

When Ag2CrO4 starts to precipitate, the [Ag+] = 3.0 x 10-5 M (see Part (a)).

When [Ag+] = 3.0 x 10-5 M,

Ksp = [Ag+][Cl-]

[Cl-] = 1.6 x 10-10

3.0 x 10-5 =5.3 x 10-6 M

15 - 27

Ion Product and PrecipitationIon Product and PrecipitationWhen mixing two solutions, how does onedetermine if a precipitate forms?

The procedure is the same as determiningthe reaction quotient, Q, in a gaseousequilibrium.

When aqueous solutions are involved, youdetermine the ion product, P.

15 - 28

The ion product of a silver chloride solution isgiven by

P = [Ag+][Cl-]

where P is compared to the Ksp of AgCl.

If P is: > Ksp, the ion concentrations are greater than the maximum for that temperature.

Precipitation occurs until P = Ksp.

15 - 29

= Ksp, the ion concentrations are a maximum for the given temperature

and equilibrium exists. < Ksp, the ion concentrations are less than the maximum and no precipitate forms.

15 - 30

Ion Product ProblemIon Product Problem0.100 L of 0.0015 M K2CrO4 is mixed with0.200 L of 0.0060 M Sr(NO3)2. Does aprecipitate form?

[K2CrO4] = 0.015 M [Sr(NO3)2] = 0.0060 M

V1 = 0.100 L V2 = 0.200 L[K2CrO4] = n

V

n = 0.015 mol K2CrO4

L× 0.100 L

15 - 31

n = 1.5 x 10-3 mol K2CrO4

n = 1.2 x 10-3 mol Sr(NO3)2

n =

× 0.200 L0.0060 mol Sr(NO3)2L

[Sr(NO3)2] = nV

= 1.2 x 10-3 mol Sr(NO3)20.300 L

15 - 32

.[Sr(NO3)2] = 4.0 x 10-3 M

[K2CrO4] = 1.5 x 10-3 mol K2CrO40.300 L

[K2CrO4] =5.0 x 10-3 M

15 - 33

.

SrCrO4(s) Sr2+(aq) + CrO42-(aq)

[ ]i 0 0[ ]c +4.0 x 10-3 +5.0 x 10-

3

[ ]e 4.0 x 10-3 5.0 x 10-3

P = [Sr2+][CrO42-]

P = 4.0 x 10-3 × 5.0 x 10-3 = 2.0 x 10-5

P = 2.0 x 10-5 < Ksp(SrCrO4) = 3.6 x 10-5

No precipitate will form.

15 - 34

Life After A PrecipitateLife After A Precipitate35.0 mL of 0.150 M Pb(NO3)2 is mixed with15.0 mL of 0.800 M KIO3.

[Pb(NO3)2] = 0.150 M [KIO3] = 0.800 M

V1 = 35.0 mL V2 = 15.0 mLKsp(Pb(IO3)2) = 2.6 x 10-13

(a)Does a precipitate form?

[Pb(NO3)2]c × Vc = [Pb(NO3)2]d × Vd

15 - 35

.

[Pb(NO3)2]d = 0.150 M × 35.0 mL50.0 mL = 0.105 M

[KIO3]d = 0.800 M × 15.0 mL50.0 mL = 0.240 M

Pb(IO3)2(s) Pb2+(aq) + 2IO3-(aq)

P = [Pb2+] [ IO3-]2 =0.105 × 0.2402

P = 6.0 x 10-3

15 - 36

P = 6.0 x 10-3 > Ksp = 2.6 x 10-13, therefore aprecipitate forms.

(b) If a precipitate forms, what are the concentrations of Pb2+ and IO3

- in the final solution?

This part is a little tricky because your first

thought might be to think the initial concentrations of Pb2+ and IO3

2- will be zero, use the Ksp, and determine the concentrations from that calculation.

15 - 37

However, P > Ksp, meaning that the concentrations of Pb2+ (0.105 M) andIO3

2- (0.240 M) are too large.

Before equilibrium can be reestablished,there is a small window in which astoichiometry (limiting reactant) scenarioexists.

A precipitate forms, meaning

nb 0.105 0.240Pb2+(aq) + 2IO3

-(aq) → Pb(IO3)2(s)na 0 0.030

15 - 38

This stoichiometry is very important becauseit says that when the shift occurs toreestablish the equilibrium, [Pb2+] = 0 and[IO3

-] = 0.030 M.

PbIO3(s) Pb2+(aq) + IO32-(aq)

[ ]i 0 0.030[ ]c +x +2x[ ]e x 0.030 + 2x

Ksp = [Pb2+][IO32-]

15 - 39

2.6 x 10-13 = x

0.030 + 2x ≈ 0.030

[Pb2+] = 2.9 x 10-10 M

[IO32-] = 0.030 M + 2 2.9 x 10-10 M

× (0.030 + 2x)2

× ≈ 0.030 M

15 - 40

pH and SolubilitypH and SolubilityThe pH or pOH of a solution can have adrastic affect on the solubility of a salt.

If HCl is added to a solution of the slightlysoluble salt, CaCO3, the solubility of the salt

isincreased.

This is an example of Le Chatelier’s principle. Before adding HCl, the following equilibrium exists:

15 - 41

CaCO3(s) Ca2+(aq) + CO32-(aq)

Adding HCl results in the ionization of HCl.

HCl(aq) → H+(aq) + Cl-(aq) CO3

2- is the conjugate base of carbonic acid which decreases the concentration of CO3

2- as a result of the following reaction:

CaCO3(s) + 2H+(aq) → Ba2+(aq) + CO2(g) + H2O(l)

15 - 42

The system will respond to the stress brought about by the lowering of the [CO3

2-] by dissolving more of the slightly soluble salt.

A similar situation arises when HCl is addedto a solution of the slightly soluble salt, Fe(OH)3.

In this case, the [OH-] is decreased because of the formation of H2O as shown below:

H+(aq) + OH-(aq) → H2O(l)

15 - 43

A similar situation arises when KOH is added

to a solution of the slightly soluble salt, Fe(OH)3.

In this case, the [OH-] is increased by the

addition of KOH which suppresses the dissolving of Fe(OH)3.

15 - 44

pH and Solubility ProblempH and Solubility Problem(a)Determine the solubility of iron(III) hydroxide in pure water. Ksp = 4 x 10-38 Fe(OH)3(s) Fe3+(aq) + 3OH-(aq)[ ]i 0 1.0 x 10-7

[ ]c+x +3x[ ]e x 1.0 x 10-7 + 3x

Ksp = [Fe3+][OH-]3

4 x 10-38 = x 27x3

x = s = 4 x 10-17 M

×

15 - 45

Some observations from the previousproblem:

Considering the Ksp is minuscule, it is absolutely necessary to consider the [OH-] found in pure water. The approximation (1.0 x 10-7 + 3x) ≈ 1.0 x 10-7 is an excellent

approximation when considering the Ksp value.

15 - 46

(b) Determine the solubility of iron(III) hydroxide in an aqueous solution whose

pH = 5.0.

pH + pOH = 14.00 pOH = 9.0 [OH-] = 10-9

Fe(OH)3(s) Fe3+(aq) + 3OH-

(aq)[ ]i 0 10-9

[ ]c +x +3x[ ]e x 10-9 + 3x

15 - 47

Ksp = [Fe3+][OH-]3

4 x 10-38 = x (10-9 + 3x)3 ≈ 10-27xx = s = 4 x 10-11 M

Unless told otherwise, assume that there isno change in the pH.

×

15 - 48

(c) Determine the solubility of iron(III) hydroxide in an aqueous solution whose

pH = 11.0.

pH + pOH = 14.00 pOH = 3.0 [OH-] = 10-3

Fe(OH)3(s) Fe3+(aq) + 3OH-

(aq)[ ]i 0 10-3

[ ]c +x +3x[ ]e x 10-3 + 3x

15 - 49

Ksp = [Fe3+][OH-]3

4 x 10-38 = x (10-3 + 3x)3 ≈ 10-9xx = s = 4 x 10-29 M

Unless told otherwise, assume that there isno change in the pH.

×

15 - 50

In conclusion:

In pure water, s = 4 x 10-17 M. This very small solubility is not surprising when considering the Ksp.

In an acidic solution, s = 4 x 10-11 M. This drastic increase is not surprising when considering Le Chatelier’s principle.

15 - 51

The stress on the system results from the formation of water which

decreases the [OH-].

The equilibrium shifts to the right resulting in an increased solubility.

In a basic solution, s = 4 x 10-29 M. Because of the large [OH-] resulting

from the pH of the solution, the dissolving of the salt will be suppressed.