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11 Measuring K[sub(sp)] and Calculating Solubility from K[sub(sp)]

# 11 Measuring K[sub(sp)] and Calculating Solubility from K[sub(sp)]

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15.11 MEASURING KSP AND CALCULATING SOLUBILITY FROM KSP

of solid CaF2 and a saturated solution of CaF2. Suppose, for example, that we mix

solutions of CaCl2 and NaF, allow time for equilibrium to be reached, and then

measure [Ca2+] = 3.5 * 10-5 M and [F -] = 1.0 * 10-3 M. Within experimental

error, these ion concentrations yield the same value of Ksp:

Ksp = [Ca2 + ][F -]2 = (3.5 * 10-5)(1.0 * 10-3)2 = 3.5 * 10-11

The value of Ksp is unaffected by the presence of other ions in solution, such as

Na+ from NaF and Cl - from CaCl2, as long as the solution is very dilute. As ion concentrations increase, Ksp values are somewhat modified because of electrostatic

interactions between ions, but we’ll ignore that complication here.

If the saturated solution is prepared by a method other than the dissolution of

CaF2 in pure water, there are no separate restrictions on [Ca2+] and [F -]; the only

restriction on the ion concentrations is that the value of the equilibrium constant

expression [Ca2+][F -]2 must equal the Ksp. That condition is satisfied by an infinite

number of combinations of [Ca2+] and [F -], and therefore we can prepare many different solutions that are saturated with respect to CaF2. For example, if [F -] is

1.0 * 10-2 M, then [Ca2+] must be 3.5 * 10-7 M:

[Ca2 + ] =

Ksp

[F -]2

=

3.5 * 10-11

(1.0 * 10-2) 2

= 3.5 * 10-7

Selected values of Ksp for various ionic compounds at 25 °C are listed in Table

15.2, and additional values can be found in Appendix C. Like all equilibrium constants, values of Ksp depend on temperature (Section 13.9).

TABLE 15.2

Ksp Values for Some Ionic Compounds at 25 °C

Name

Formula

Aluminum hydroxide

Al(OH)3

Barium carbonate

Calcium carbonate

Calcium fluoride

Silver chloride

Silver sulfate

BaCO3

CaCO3

CaF2

PbCl2

PbCrO4

AgCl

Ag2SO4

Ksp

1.9

2.6

5.0

3.5

1.2

2.8

1.8

1.2

*

*

*

*

*

*

*

*

10-33

10-9

10-9

10-11

10-5

10-13

10-10

10-5

Once the Ksp value for a compound has been measured, you can use it to calculate the solubility of the compound—the amount of compound that dissolves per

unit volume of saturated solution. Because of two complications, however, calculated solubilities are often approximate. First, Ksp values can be difficult to measure,

and values listed in different sources might differ by as much as a factor of 10 or

more. Second, calculated solubilities can be less than observed solubilities because of

side reactions. For example, dissolution of PbCl2 gives both Pb 2+ and PbCl + because

of some ion association between Pb 2+ and Cl - ions:

(1) PbCl2(s) Δ Pb2 + (aq) + 2 Cl -(aq)

(2) Pb 2 + (aq) + Cl -(aq) Δ PbCl +(aq)

In this book, we will calculate approximate solubilities assuming that ionic solutes

are completely dissociated (reaction 1). In the case of PbCl2, ignoring the second

equilibrium gives a calculated solubility that is too low by a factor of about 2.

613

614

Chapter 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

WORKED EXAMPLE 15.8

CALCULATING Ksp FROM ION CONCENTRATIONS

A particular saturated solution of silver chromate, Ag2CrO4, has [Ag +] = 5.0 * 10-5 M

and [CrO 4 2-] = 4.4 * 10-4 M. What is the value of Ksp for Ag2CrO4?

STRATEGY

Substituting the equilibrium concentrations into the expression for Ksp of Ag2CrO4

(Worked Example 15.7) gives the value of Ksp.

SOLUTION

Ksp = [Ag +]2[CrO 4 2-] = (5.0 * 10-5)2(4.4 * 10-4) = 1.1 * 10-12

BALLPARK CHECK

Ksp is approximately (5)2(4) = 100 times (10-5)2(10-4). So Ksp is about 10-12, in agreement with the solution.

WORKED EXAMPLE 15.9

᭡ Addition of aqueous K2CrO4 to aqueous

AgNO3 gives a red precipitate of Ag2CrO4

and a saturated solution of Ag2CrO4.

CALCULATING Ksp FROM SOLUBILITY

A saturated solution of Ag2CrO4 prepared by dissolving solid Ag2CrO4 in water has

[CrO 4 2-] = 6.5 * 10-5 M. Calculate Ksp for Ag2CrO4.

STRATEGY

Because both the Ag + and CrO 4 2- ions come from the dissolution of solid Ag2CrO4,

[Ag +] must be twice [CrO 4 2-].

Ag2CrO4(s) Δ 2 Ag +(aq) + CrO 4 2-(aq)

Substituting [Ag +] and [CrO 4 2-] into the equilibrium equation gives the value of Ksp.

SOLUTION

[Ag +] = (2)[CrO 4 2-] = (2)(6.5 * 10-5) = 1.3 * 10-4 M

Ksp = [Ag +]2[CrO 4 2-] = (1.3 * 10-4)2(6.5 * 10-5) = 1.1 * 10-12

WORKED EXAMPLE 15.10

CALCULATING SOLUBILITY FROM KSP

Calculate the solubility of MgF2 in water at 25 °C in units of:

(a) Moles per liter

(b) Grams per liter

STRATEGY

(a) Write the balanced equation for the solubility equilibrium assuming the complete

dissociation of MgF2, and look up Ksp for MgF2 in Appendix C. If we define x as the

number of moles per liter of MgF2 that dissolves, then the saturated solution contains

x mol/L of Mg 2+ and 2x mol/L of F -. Substituting these equilibrium concentrations

into the expression for Ksp and solving for x gives the molar solubility.

(b) To convert the solubility from units of moles per liter to units of grams per liter,

multiply the molar solubility of MgF2 by its molar mass (62.3 g/mol).

SOLUTION

(a) It’s helpful to summarize the equilibrium concentrations under the balanced equation:

Solubility Equilibrium

Equilibrium concentration (M)

MgF2(s) Δ Mg 2؉(aq) ؉ 2 F ؊ (aq)

x

2x

15.11 MEASURING KSP AND CALCULATING SOLUBILITY FROM KSP

Substituting the equilibrium concentrations into the expression for Ksp gives

Ksp = 7.4 * 10-11 = [Mg 2 + ][F -]2 = (x)(2x)2

4x 3 = 7.4 * 10-11

x 3 = 1.8 * 10-11

x = [Mg 2 + ] = Molar solubility = 2.6 * 10-4 mol/L

Thus, the molar solubility of MgF2 in water at 25 °C is 2.6 * 10-4 M. (Taking

account of the side reaction that produces MgF + would increase the calculated solubility by about 6%.)

Note that the number 2 appears twice in the expression (x)(2x)2. The exponent 2

is required because of the equilibrium equation, Ksp = [Mg 2+][F -]2. The coefficient

2 in 2x is required because each mole of MgF2 that dissolves gives 2 mol of F -(aq).

(b) Solubility (in g/L) =

62.3 g

2.6 * 10-4 mol

= 1.6 * 10-2g/L

*

L

mol

BALLPARK CHECK

(a) The molar solubility of MgF2 is the cube root of Ksp/4, which equals the cube root of

approximately 20 * 10-12. Because (2)3 = 8, (3)3 = 27, and (10-4)3 = 10-12, the molar

solubility is between 2 * 10-4 M and 3 * 10-4 M. The ballpark check and the solution

agree.

A saturated solution of Ca3(PO4)2 has [Ca2+] = 2.01 * 10-8 M and

] = 1.6 * 10 M. Calculate Ksp for Ca3(PO4)2.

Ī PROBLEM 15.21

[PO 4

3-

-5

Ī PROBLEM 15.22 Prior to having an X-ray exam of the upper gastrointestinal tract, a

patient drinks an aqueous suspension of solid BaSO4. (Scattering of X rays by barium

greatly enhances the quality of the photograph.) Although Ba2+ is toxic, ingestion of BaSO4

is safe because it is quite insoluble. If a saturated solution prepared by dissolving solid

BaSO4 in water has [Ba2+] = 1.05 * 10-5 M, what is the value of Ksp for BaSO4?

Ī PROBLEM 15.23

or Ag2CrO4 with Ksp

Which has the greater molar solubility: AgCl with Ksp = 1.8 * 10-10

= 1.1 * 10-12? Which has the greater solubility in grams per liter?

CONCEPTUAL PROBLEM 15.24 The following pictures represent saturated solutions

of three silver salts: AgX, AgY, and AgZ. (Other ions and solvent water molecules have

been omitted for clarity.)

= Ag+

AgX

= X–, Y–, or Z–

AgY

(a) Which salt has the largest value of Ksp?

(b) Which salt has the smallest value of Ksp?

AgZ

᭡ This X-ray photograph of the

small intestine was taken soon

after the patient drank a

barium sulfate “cocktail.”

615

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Chapter 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

15.12 FACTORS THAT AFFECT SOLUBILITY

Solubility and the Common-Ion Effect

We’ve already discussed the common-ion effect in connection with the dissociation

of weak acids and bases (Section 15.2). To see how a common ion affects the position

of a solubility equilibrium, let’s look again at the solubility of MgF2:

MgF2(s) Δ Mg 2 + (aq) + 2 F -(aq)

In Worked Example 15.10, we found that the molar solubility of MgF2 in pure water

at 25 °C is 2.6 * 10-4 M. Thus,

[Mg 2 + ] = 2.6 * 10-4 M

[F -] = 5.2 * 10-4 M

When MgF2 dissolves in a solution that contains a common ion from another

source—say, F - from NaF—the position of the solubility equilibrium is shifted to the

left by the common-ion effect. If [F -] is larger than 5.2 * 10-4 M, then [Mg 2+] must

be correspondingly smaller than 2.6 * 10-4 M to maintain the equilibrium expression [Mg 2+][F -]2 at a constant value of Ksp = 7.4 * 10-11. A smaller value of [Mg 2+]

means that MgF2 is less soluble in a sodium fluoride solution than it is in pure water.

Similarly, the presence of Mg 2+ from another source—say, MgCl2—shifts the solubility equilibrium to the left and decreases the solubility of MgF2.

. . . if this increases . . .

This must decrease . . .

. . . because this is constant.

[Mg2+][F−]2 = Ksp = 7.4 × 10−11

If this increases . . .

. . . because this is constant.

. . . this must decrease . . .

Solubility of MgF2 (mol/L)

In general, the solubility of a slightly soluble ionic compound is decreased by the

presence of a common ion in the solution, as illustrated in Figure 15.12. The quantitative aspects of the common-ion effect are explored in Worked Example 15.11.

Figure 15.12

Pure water

10–4

The solubility of MgF2

at 25 °C decreases

markedly on the

10–6

10–8

0

The common-ion effect. The calculated

solubility of MgF2 is plotted on a

logarithmic scale.

0.040

0.080

0.120

WORKED EXAMPLE 15.11

CALCULATING SOLUBILITY IN A SOLUTION

THAT CONTAINS A COMMON ION

Calculate the molar solubility of MgF2 in 0.10 M NaF at 25 °C.

STRATEGY

Once again, we define x as the molar solubility of MgF2. The dissolution of x mol/L of

MgF2 provides x mol/L of Mg 2+ and 2x mol/L of F - , but the total concentration of F -

15.12 FACTORS THAT AFFECT SOLUBILITY

617

is (0.10 + 2x) mol/L because the solution already contains 0.10 mol/L of F - from the

completely dissociated NaF. Substituting the equilibrium concentrations into the

expression for Ksp and solving for x gives the molar solubility.

SOLUTION

Again, we summarize the equilibrium concentrations under the balanced equation:

MgF2(s) Δ Mg 2؉ (aq) + 2 F ؊ (aq)

Solubility Equilibrium

0.10 + 2x

x

Equilibrium concentration (M)

Substituting the equilibrium concentrations into the expression for Ksp gives

Ksp = 7.4 * 10-11 = [Mg 2+][F -]2 = (x)(0.10 + 2x)2

Because Ksp is small, 2x will be small compared to 0.10 and we can make the approximation that (0.10 + 2x) L 0.10. Therefore,

7.4 * 10-11 = (x)(0.10 + 2x)2 L (x)(0.10)2

x = [Mg 2 + ] = Molar solubility =

7.4 * 10-11

(0.10)2

= 7.4 * 10-9 M

Note that the calculated solubility of MgF2 in 0.10 M NaF is less than that in pure water

by a factor of about 35,000! (See Figure 15.12.) (Taking account of the equilibrium

Mg 2+ + F - Δ MgF + would increase the calculated solubility by a factor of about 10.

In this case, quite a bit of the Mg 2+ is converted to MgF + because of the relatively high

F - concentration in 0.10 M NaF.)

BALLPARK CHECK

We can check our results by substituting the calculated equilibrium concentrations into

the expression for Ksp:

Ksp = 7.4 * 10-11 = [Mg 2 + ][F -]2 = (7.4 * 10-9)(0.10)2 = 7.4 * 10-11

Ī PROBLEM 15.25

Calculate the molar solubility of MgF2 in 0.10 M MgCl2 at 25 °C.

Solubility and the pH of the Solution

Solubility of CaCO3 (mol/L)

An ionic compound that contains a basic anion becomes more soluble as the acidity of

the solution increases. The solubility of CaCO3, for example, increases with decreasing pH (Figure 15.13) because the CO3 2- ions combine with protons to give HCO3 ions. As CO3 2- ions are removed from the solution, the solubility equilibrium shifts to

10–1

The solubility of CaCO3 increases

as the [H3O+] increases because

the CO32− ions combine with

protons, thus driving the

solubility equilibrium

to the right.

10–2

10–3

10–4

Figure 15.13

4

6

8

10

pH

12

14

Plot of the solubility of CaCO3 at 25 °C

versus the pH of the solution. The

solubility is plotted on a logarithmic

scale.

618

Chapter 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

the right, as predicted by Le Châtelier’s principle. The net reaction is the dissolution

of CaCO3 in acidic solution to give Ca2+ ions and HCO3 - ions:

CaCO3(s) Δ Ca2 + (aq) + CO3 2-(aq)

H 3O +(aq) + CO3 2-(aq) Δ HCO3 -(aq) + H 2O(l)

Net:

᭡ These downward-growing, icicleshaped structures, called stalactites, and

the upward growing columns, called

stalagmites, are formed in limestone

caves by the slow precipitation of

calcium carbonate from dripping water.

CaCO3(s) + H 3O +(aq) Δ Ca2 + (aq) + HCO3 -(aq) + H 2O(l)

Other salts that contain basic anions, such as CN - , PO4 3- , S 2- , or F - , behave similarly. By contrast, pH has no effect on the solubility of salts that contain anions of

strong acids (Cl - , Br - , I - , NO3 - , and ClO4 -) because these anions are not protonated

by H 3O +.

The effect of pH on the solubility of CaCO3 has important environmental consequences. For instance, the formation of limestone caves, such as Mammoth Cave in

Kentucky, is due to the slow dissolution of limestone (CaCO3) in the slightly acidic

natural water of underground streams, and the formation of stalactites and stalagmites in these caves is due to the slow precipitation of CaCO3 from dripping water.

Marble, another form of CaCO3, also dissolves in acid, which accounts for the deterioration of marble monuments on exposure to acid rain (Inquiry, pages 576–577).

The effect of pH on solubility is also important in understanding how fluoride

ion reduces tooth decay. When tooth enamel, mainly hydroxyapatite, Ca5(PO4)3OH,

comes in contact with F - ions in drinking water or fluoride-containing toothpaste,

OH - ions in Ca5(PO4)3OH are replaced by F - ions, giving the mineral fluorapatite,

Ca5(PO4)3F. Because F - is a much weaker base than OH - , Ca5(PO4)3F is much more

resistant than Ca5(PO4)3OH to dissolving in acids. (See the Inquiry at the end of this

chapter.)

Ī PROBLEM 15.26 Which of the following compounds are more soluble in acidic solution than in pure water?

(a) AgCN

(b) PbI2

(c) Al(OH)3

(d) ZnS

Solubility and the Formation of Complex Ions

Remember ...

A coordinate covalent bond forms when

one atom donates two electrons (a lone

pair) to another atom that has a vacant

orbital. (Section 7.5)

The electron-pair donor is a Lewis base.

(Section 14.16)

The solubility of an ionic compound increases dramatically if the solution contains a

Lewis base that can form a coordinate covalent bond to the metal cation (Section 7.5).

Silver chloride, for example, is quite insoluble in water and in acid, but it dissolves in

an excess of aqueous ammonia, forming the complex ion Ag(NH 3)2 + (Figure 15.14). A

complex ion is an ion that contains a metal cation bonded to one or more small molecules or ions, such as NH3, CN -, or OH -. In accord with Le Châtelier’s principle,

ammonia shifts the solubility equilibrium to the right by tying up the Ag + ion in the

form of the complex ion:

AgCl(s) Δ Ag +(aq) + Cl -(aq)

Ag +(aq) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq)

The formation of a complex ion is a stepwise process, and each step has its own

characteristic equilibrium constant. For the formation of Ag(NH 3)2 +, the reactions are

Ag +(aq) + NH 3(aq) Δ Ag(NH 3) + (aq)

K1 = 2.1 * 103

Ag(NH 3)+(aq) + NH 3(aq) Δ Ag(NH 3)2 +(aq)

K2 = 8.1 * 103

Net: Ag +(aq) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq)

K f = 1.7 * 107

K1 =

[Ag(NH 3)+]

[Ag +][NH 3]

= 2.1 * 10

Kf = K1K2 =

3

[Ag(NH 3)2 +]

[Ag +][NH 3]2

K2 =

[Ag(NH 3)2 +]

[Ag(NH 3) +][NH 3]

= 1.7 * 107

at 25 °C

= 8.1 * 103

15.12 FACTORS THAT AFFECT SOLUBILITY

Figure 15.14

Silver chloride is quite insoluble in

water . . .

. . . but dissolves on addition of an

excess of aqueous ammonia.

Cl–

Cl–

Ag+

NH3

NH3

Ag(NH3)2+

The stability of a complex ion is measured by its formation constant Kf (or

stability constant), the equilibrium constant for the formation of the complex ion from

the hydrated metal cation. The large value of Kf for Ag(NH 3)2 + means that this complex ion is quite stable, and nearly all the Ag + ion in an aqueous ammonia solution is

therefore present in the form of Ag(NH 3)2 + (see Worked Example 15.12).

The net reaction for dissolution of AgCl in aqueous ammonia is the sum of the

equations for the dissolution of AgCl in water and the reaction of Ag +(aq) with

NH 3(aq) to give Ag(NH 3)2 + :

AgCl(s) Δ Ag +(aq) + Cl -(aq)

Ag +(aq) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq)

Net: AgCl(s) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq) + Cl -(aq)

The equilibrium constant K for the net reaction is the product of the equilibrium

K =

[Ag(NH 3)2 +][Cl -]

[NH 3]2

= (Ksp)(Kf) = (1.8 * 10-10)(1.7 * 107) = 3.1 * 10-3

Because K is much larger than Ksp, the solubility equilibrium for AgCl lies

much farther to the right in the presence of ammonia than it does in the absence of

ammonia. The increase in the solubility of AgCl on the addition of ammonia is

shown graphically in Figure 15.15. In general, the solubility of an ionic compound

increases when the metal cation is tied up in the form of a complex ion. The quantitative effect of complex formation on the solubility of AgCl is explored in Worked

Example 15.13.

The reaction of silver chloride with

aqueous ammonia.

619

620

Chapter 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

Figure 15.15

Solubility of AgCl (mol/L)

Plot of the solubility of AgCl in

aqueous ammonia at 25 °C versus the

concentration of ammonia. The

solubility is plotted on a logarithmic

scale.

10–1

10–2

The solubility of AgCl in

NH3(aq) increases with increasing

[NH3] owing to formation of the

complex ion Ag(NH3)2+.

10–3

10–4

Pure water

10–5

0

1.0

2.0

3.0

Concentration of NH3 (mol/L)

WORKED EXAMPLE 15.12

CALCULATING THE CONCENTRATIONS OF COMPLEX IONS

What are the concentrations of Ag + , Ag(NH 3)+ , and Ag(NH 3)2 + in a solution prepared

by adding 0.10 mol of AgNO3 to 1.0 L of 3.0 M NH3? Kf = 1.7 * 107, K1 = 2.1 * 103,

and K2 = 8.1 * 103.

STRATEGY

Because K1, K2, and Kf for Ag(NH 3)2 + are all large numbers, nearly all the Ag + from

AgNO3 will be converted to Ag(NH 3)2 + :

Ag +(aq) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq)

Kf = 1.7 * 107

To calculate the concentrations, it’s convenient to imagine that 100% of the Ag + is converted to Ag(NH 3)2 + , followed by a tiny amount of back-reaction [dissociation of

Ag(NH 3)2 +] to give a small equilibrium concentration of Ag + .

SOLUTION

The conversion of 0.10 mol/L of Ag + to Ag(NH 3)2 + consumes 0.20 mol/L of NH3.

Assuming 100% conversion to Ag(NH 3)2 + , the following concentrations are obtained:

[Ag +] = 0 M

[Ag(NH 3)2 +] = 0.10 M

[NH 3] = 3.0 - 0.20 = 2.8 M

The dissociation of x mol/L of Ag(NH 3)2 + in the back-reaction produces x mol/L of

Ag + and 2x mol/L of NH3. Therefore, the equilibrium concentrations (in mol/L) are

[Ag(NH 3)2 +] = 0.10 - x

[Ag +] = x

[NH 3] = 2.8 + 2x

Let’s summarize our reasoning in a table under the balanced equation:

Ag ؉ (aq) ؉ 2 NH3(aq) Δ Ag(NH3)2؉ (aq)

Initial concentration (M)

0.10

3.0

0

After 100% reaction (M)

0

2.8

0.10

Equilibrium concentration (M)

x

2.8 + 2x

0.10 - x

15.12 FACTORS THAT AFFECT SOLUBILITY

Substituting the equilibrium concentrations into the expression for Kf and making

the approximation that x is negligible compared to 0.10 (and to 2.8) gives

Kf = 1.7 * 107 =

[Ag +] = x =

[Ag(NH 3)2 +]

+

2

[Ag ][NH 3]

0.10

=

0.10 - x

L

2

(x)(2.8 + 2x)

0.10

(x)(2.8)2

= 7.5 * 10-10 M

(1.7 * 107)(2.8)2

[Ag(NH 3)2 +] = 0.10 - x = 0.10 - (7.5 * 10-10) = 0.10 M

The concentration of Ag(NH 3)+ can be calculated from either of the stepwise equilibria. Let’s use the equilibrium equation for the formation of Ag(NH 3)+ from Ag + :

K1 =

[Ag(NH 3)+]

[Ag +][NH 3]

= 2.1 * 103

[Ag(NH 3)+] = K1[Ag +][NH 3] = (2.1 * 103)(7.5 * 10-10)(2.8) = 4.4 * 10-6 M

Thus, nearly all the Ag + is in the form of Ag(NH 3)2 + .

BALLPARK CHECK

The approximate equilibrium concentrations are [Ag +] = 7 * 10-10 M, [Ag(NH 3)+] =

4 * 10-6 M, [Ag(NH 3)2 +] = 0.1 M, and [NH 3] = 3 M. We can check these results by

substituting them into the equilibrium constant expressions for Kf and K1:

Kf = 1.7 * 107 =

K1 = 2.1 * 103 =

[Ag(NH 3)2 +]

+

2

[Ag ][NH 3]

[Ag(NH 3)+]

[Ag +][NH 3]

L

L

0.1

(7 * 10-10)(3)2

4 * 10-6

(7 * 10-10)(3)

= 2 * 107

= 2 * 103

The ballpark estimates of Kf and K1 agree with the experimental values.

WORKED EXAMPLE 15.13

EXPLORING THE EFFECT OF COMPLEX FORMATION ON SOLUBILITY

Calculate the molar solubility of AgCl at 25 °C in:

(a) Pure water

(b) 3.0 M NH3

STRATEGY

Write the balanced equation for the dissolution reaction, and define x as the number of

moles per liter of AgCl that dissolves. Then express the equilibrium concentrations in

terms of x, and substitute them into the appropriate equilibrium equation. Solving for x

gives the molar solubility.

SOLUTION

(a) In pure water, the solubility equilibrium is

AgCl(s) Δ Ag +(aq) + Cl -(aq)

Substituting the equilibrium concentrations (x mol/L) into the expression for Ksp

gives

Ksp = 1.8 * 10-10 = [Ag +][Cl -] = (x)(x)

x = Molar solubility = 21.8 * 10-10 = 1.3 * 10-5 M

(b) The balanced equation for the dissolution of AgCl in aqueous NH3 is

AgCl(s) + 2 NH 3(aq) Δ Ag(NH 3)2 +(aq) + Cl -(aq)

K = 3.1 * 10-3

If we define x as the number of moles per liter of AgCl that dissolves, then the

saturated solution contains x mol/L of Ag(NH 3)2 + , x mol/L of Cl - , and

(3.0 - 2x) mol/L of NH3. (We’re assuming that essentially all the Ag + is in the form

continued on next page

621

622

Chapter 15 APPLICATIONS OF AQUEOUS EQUILIBRIA

of Ag(NH 3)2 + , as proved in Worked Example 15.12.) Substituting the equilibrium

concentrations into the equilibrium equation gives

[Ag(NH 3)2 +][Cl -]

(x)(x)

=

K = 3.1 * 10-3 =

2

[NH 3]

(3.0 - 2x)2

Taking the square root of both sides, we obtain

x

5.6 * 10-2 =

3.0 - 2x

x = (5.6 * 10-2)(3.0 - 2x) = 0.17 - 0.11x

0.17

x =

= 0.15 M

1.11

The molar solubility of AgCl in 3.0 M NH3 is 0.15 M. Thus, AgCl is much more

soluble in aqueous NH3 than in pure water, as shown in Figure 15.15.

BALLPARK CHECK

Check the calculated equilibrium concentrations by substituting them into the appropriate equilibrium equation. For part (b), for example, [Ag(NH 3)2 +] = [Cl -] = x =

0.15 M and [NH 3] = 3.0 - 2x = 2.7 M. Since [NH3] is 10% less than 3 M, K =

[Ag(NH3)2+][Cl-]/[NH3]2 is about 20% greater than (0.15/3)2 = (0.05)2 = 2.5 * 10-3.

The ballpark estimate and the experimental value of K = 3.1 * 10-3 agree.

In an excess of NH 3(aq), Cu2+ ion forms a deep blue complex ion,

Cu(NH 3)4 , which has a formation constant Kf = 5.6 * 1011. Calculate the concentration

of Cu2+ in a solution prepared by adding 5.0 * 10-3 mol of CuSO4 to 0.500 L of 0.40 M NH3.

Ī PROBLEM 15.27

2+

Ī PROBLEM 15.28 Silver bromide dissolves in aqueous sodium thiosulfate, Na2S2O3,

yielding the complex ion Ag(S 2O3)2 3-:

AgBr(s) + 2 S 2O3 2-(aq) Δ Ag(S 2O3)2 3-(aq) + Br -(aq)

K = ?

Calculate the equilibrium constant K for the dissolution reaction, and calculate the molar

solubility of AgBr in 0.10 M Na2S2O3. Ksp = 5.4 * 10-13 for AgBr and Kf = 4.7 * 1013

for Ag(S 2O3)2 3- .

Solubility and Amphoterism

Certain metal hydroxides, such as aluminum hydroxide (Figure 15.16), are soluble

both in strongly acidic and in strongly basic solutions:

In acid: Al(OH)3(s) + 3 H 3O +(aq) Δ Al3+(aq) + 6 H 2O(l)

In base: Al(OH)3(s) + OH -(aq) Δ Al(OH)4 -(aq)

Figure 15.16

Such hydroxides are said to be amphoteric (am-fo-tare-ic), a term that comes from

the Greek word amphoteros, meaning “in both ways.”

The amphoteric behavior of Al(OH)3.

Aluminum hydroxide,

a gelatinous white precipitate,

NaOH to Al3+(aq).

The precipitate dissolves on

NaOH, yielding the colorless

Al(OH)4– ion. The precipitate

also dissolves in aqueous HCl,

yielding the colorless Al3+ ion.

15.13 PRECIPITATION OF IONIC COMPOUNDS

623

The dissolution of Al(OH)3 in excess base is just a special case of the effect of

complex-ion formation on solubility: Al(OH)3 dissolves because excess OH - ions

convert it to the soluble complex ion Al(OH)4 - (aluminate ion). The effect of pH on

the solubility of Al(OH)3 is shown in Figure 15.17.

Figure 15.17

Solubility of Al(OH)3 (mol/L)

Al(OH)3 dissolves in

strongly acidic solutions.

Al(OH)3 dissolves in

strongly basic solutions.

A plot of solubility versus pH shows

that Al(OH)3 is an amphoteric

hydroxide.

1.00

0.75

0.50

Al(OH)3 is essentially

insoluble between

pH 4 and 10.

0.25

0

2

4

6

8

10

12

pH

Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)3, Sn(OH)2,

and Pb(OH)2, which react with excess OH - ions to form the soluble complex ions

Zn(OH)4 2- (zincate ion), Cr(OH)4 - (chromite ion), Sn(OH)3 - (stannite ion), and

Pb(OH)3 - (plumbite ion), respectively. By contrast, basic hydroxides, such as

Mn(OH)2, Fe(OH)2, and Fe(OH)3, dissolve in strong acid but not in strong base.

15.13 PRECIPITATION OF IONIC COMPOUNDS

A common problem in chemistry is to decide whether a precipitate of an ionic compound will form when solutions that contain the constituent ions are mixed. For

example, will CaF2 precipitate on mixing solutions of CaCl2 and NaF? In other

words, will the dissolution reaction proceed in the reverse direction, from right

to left?

CaF2(s) Δ Ca2 + (aq) + 2 F -(aq)

We touched on this question briefly in Section 4.4 when we looked at solubility

guidelines, but we can now get a more quantitative view. The answer depends on the

value of the ion product (IP), a number defined by the expression

IP = [Ca2 + ]t[F -]t 2

The IP is defined in the same way as Ksp, except that the concentrations in the

expression for IP are initial concentrations—that is, arbitrary concentrations at time t,

not necessarily equilibrium concentrations. Thus, the IP is actually a reaction

quotient Qc (Section 13.5), but the term ion product is more descriptive because, as

usual, solid CaF2 is omitted from the equilibrium constant expression.

If the value of IP is greater than Ksp, the solution is supersaturated with respect to

CaF2—a nonequilibrium situation. In that case, CaF2 will precipitate, thus reducing

the ion concentrations until IP equals Ksp. At that point, solubility equilibrium is

reached and the solution is saturated.

Remember ...

The reaction quotient Qc is defined in the

same way as the equilibrium constant Kc

except that the concentrations in the equilibrium constant expression are not necessarily

equilibrium values. (Section 13.5)

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