Practice Problems Acid-Base Equilibria and Solubility Equilibria

1. Write the solubility product constant expression for equilibrium in a saturated solution of (a) iron(III) hydroxide and (b) gold(III) oxalate.

Fe(OH)3(s) → ← Fe3+(aq) + 3 OH–(aq)

Au2(C2O4)3(s) → ← 2 Au3+(aq) + 3 C2O42–(aq)

Answer

Solids do not contribute to mass action expressions, so the solubility product constant is

(a) Ksp = [Fe3+]e[OH–]e3

(b) Ksp = [Au3+]e2[C2O42–]e3

2. When does pH affect the solubility of a slightly soluble solute, and when does it not? Give some examples.

Answer

The solubility of an ionic salt will be affected by pH if either the cation or the anion can undergo significant hydrolysis. This includes most salts.

Example: FePO4

FePO4(s) → ← Fe3+(aq) + PO43–(aq)

Both ions can undergo hydrolysis:

Fe3+(aq) + 2 H2O(l) → ← FeOH2+(aq) + H3O+(aq) 1 Ka = 6.3×10–3

PO43–(aq) + H2O(l) → ← HPO42–(aq) + OH–(aq) 2 Kb = 1.0×10–14/4.2×10–13 = 2.4×10–2

The base hydrolysis is slightly more important under initially neutral conditions.

At higher pH, where there is an excess of hydroxide ions, reaction (2 ) will shift towards reactants, thereby tending to reduce solubility. However, an acid/base reaction can occur:

OH–(aq) + H3O+(aq) → 2 H2O(l)

This will drive reaction (1) towards products, which will help increase solubility. The net change in solubility depends upon the pH and the balance of the various equilibria.

Likewise, at low pH, where there is an excess of hydronium ions, reaction (1) will shift towards reactants, with a tendency for reducing solubility. Again, an acid/base reaction can occur between hydroxide ions and hydronium ions which, at low pH, will drive reaction (2) towards products and increase solubility.

3. Explain why PbCl2(s) is less soluble in 1 M Pb(NO3)2 than in pure water, but somewhat more soluble in 1 M HCl(aq) than in pure water.

Answer

Look at the different reactions:

In pure water,

PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)

In 1 M Pb(NO3)2 both salts must be considered,

Pb(NO3)2(aq) → Pb2+(aq) + 2 NO3–(aq)

PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)

The common lead ion pushes the equilibrium reaction towards reactants, i.e., less solubility.

In 1 M HCl three reactions must be considered,

HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)

Pb2+(aq) + 4 Cl–(aq) → ← [PbCl4]2–(aq)

The formation of the complex ion in the third reaction drives the solubilization equilibrium of the lead(II) chloride, thereby increasing the solubility somewhat.

4. Write an equation involving [Al(H2O)6]3+ to account for the fact that aqueous solutions of Al3+ are acidic.

Answer

This is like any other cation hydrolysis if we separate one water molecule from the others:

[Al(H2O)6]3+ = [(H2O)Al(H2O)5]3+

The red hydrogen atom becomes the hydrogen ion donor in a typical Brønsted–Lowry acid:

[(H2O)Al(H2O)5]3+ (aq) + H2O(l) → ← H3O+(aq) + [Al(H2O)5(OH)]2+(aq)

5. Write a chemical equation representing solubility equilibrium for (a) Hg2(CN)2, Ksp = 5×10–40; and (b) Ag3AsO4, Ksp = 1.0 ×10–22.

Answer

(a) Hg2(CN)2(s) → ← Hg22+(aq) + 2 CN–(aq)

Ksp = [Hg22+]e[CN–]e2

(b) Ag3AsO4(s) → ← 3 Ag+(aq) + AsO43–(aq)

Ksp = [Ag+]e3[AsO43–]e

6. Write a chemical equation representing solubility equilibrium for (a) YF3; Ksp = 6.6×10–13; and (b) Fe4[Fe(CN)6]3, Ksp = 3.3×10–41.

Answer

(a) YF3(s) → ← Y3+(aq) + 3 F–(aq); Ksp = [Y3+]e[F–]e3.

(b) Fe4[Fe(CN)6]3 (s) → ← 4 Fe3+(aq) + 3 [Fe(CN)6]4–(aq); Ksp = [Fe3+]e4[Fe(CN)4–]e3.

7. The Ksp values of CuCO3 and ZnCO3 are 1.4×10–10 and 1.4×10–11, respectively. Does this mean that CuCO3 has ten times the solubility of ZnCO3? Explain.

Answer

Although the equilibrium constants differ by a factor of ten, the solubilities of each compound do not differ by a factor of ten. The solubility of a compound is determined from the value of Ksp and the stoichiometry of the ionization reaction. In this case, the solubilities of the two compounds differ by about a factor of 3.2.

8. Calculate Ksp values of the following, for which a reference book lists the indicated solubilities: (a) Ce(IO3)4, solubility 1.8×10–4 mol/L; (b) Hg2SO4, 8.9×10–4 mol/L; and (c) barium chromate, 0.0010 g BaCrO4/100 mL H2O.

Answer

Strategy: write the chemical reaction, write the mass action expression, set up a table of concentrations, convert the given solubility to molarity if necessary, and then use this value to find Ksp.

(a)

Ce(IO3)4(s) → ← Ce4+(aq) + 4 IO3–(aq)

Ksp = [Ce4+]e[IO3–]e4


Initial0 0

Change+x +4x

Equilibriumx 4x

Molar solubility of Ce(IO3)4 = x = 1.8×10–4 mol/L

[Ce4+]e = x = 1.8×10–4 mol/L

[IO3–]e = 4x = 4(1.8×10–4 mol/L) = 7.2×10–4 mol/L

Ksp = [Ce4+]e[IO3–]e4 = [1.8×10–4 mol/L][7.2×10–4 mol/L]4 = 4.8×10–17

(b)

Hg2SO4(s) → ← Hg22+(aq) + SO42–(aq)

Ksp = [Hg22+]e[SO42–]e


Initial0 0

Change+x +x

Equilibriumx x

Molar solubility of Hg2SO4 = x = 8.9×10–4 mol/L

[Hg22+]e = x = 8.9×10–4 mol/L

[SO42–]e = x = 8.9×10–4 mol/L

Ksp = [Hg22+]e[SO42–]e = [8.9×10–4 mol/L][8.9×10–4 mol/L] = 7.9×10–7

(c)

BaCrO4(s) → ← Ba2+(aq) + CrO42–(aq)

Ksp = [Ba2+]e[CrO42–]e


Initial0 0

Change+x +x

Equilibriumx x

The variable x must be found from the solubility given, after converting to molarity.

BaCrO4 = 137.3 + 52.0 + 4(16.0) = 253.3 g/mol

0.0010 g/100 mL = 0.010 g/L = 0.010 g/(253.3 g/mol)/L = 3.9×10–5 M

Since the stoichiometry between barium chromate and the ions is 1:1, x = 3.9×10–5 M.

Now this value can be used in the mass action expression for Ksp to give:

Ksp = [x][x] = [3.9×10–5] [3.9×10–5] = 1.6×10–9

9. Calculate the concentration of Cu2+ in parts per billion (ppb) in a saturated solution of copper(II) arsenate, Cu3(AsO4)2(aq). (Hint: Recall that 1 ppb signifies 1 g Cu2+ per 109 g solution.)

Answer

First, treat the solubility equilibrium in units of molarity and then change the concentration to ppb.

Cu3(AsO4)2(s) → ← 3 Cu2+(aq) + 2 AsO43–(aq)

Ksp = [Cu2+]e3[AsO43–]e2 = 7.6×10–36


Initial0 0

Change+3x +2x

Equilibrium3x 2x

7.6×10–36 = [3x]3[2x]2 = 108x5

x = 3.7×10–8 M = molar solubility of copper(II) arsenate.

The concentration of Cu2+ ions = 3x = 3(3.7×10–8) = 1.1×10–7 M.

Assuming the density of the dilute solution is about the same as the density of water, 1 g/mL = 103 g/L, then to find ppb requires 106 L of solution:

[Cu2+] = (1.1×10–7 mol/L)(63.5 g/mol)(106 L solution per 109 g solution) = 7.0 ppb

10. A solution is saturated with Ag2SO4. (a) Calculate [Ag+] in this saturated solution. (b) What mass of Na2SO4 must be added to 0.500 L of the solution to decrease [Ag+] to 4.0×10–3 M.

Answer

(a)

Ag2SO4(s) → ← 2 Ag+(aq) + SO42–(aq)

Ksp = [Ag+]e2[SO42–]e = 1.4×10–5


Initial0 0

Change+2x +x

Equilibrium2x x

1.4×10–;5 = [2x]2[x] = 4x3

x = 0.015 M = silver sulfate concentration in a saturated solution.

[Ag+] = 2x = 2(0.015) = 0.030 M.

(b)

Ag2SO4(s) → ← 2 Ag+(aq) + SO42–(aq)

Ksp = [Ag+]e2[SO42–]e = 1.4×10–5


Initial0 0

Change+2x +x

Equilibrium4.0×10–3 x

The equilibrium concentration of Ag+ is given in the problem = 4.0×10–3 M

1.4×10–5 = [4.0×10–3]2[x]

x = 0.88 M = the concentration of sulfate required to reduce the silver ion concentration to the desired level.

The number of moles required in 0.500 L = (0.88 mol/L)(0.500 L) = 0.44 mol

The molar mass of Na2SO4 = 2(23.0) + 32.1 + 4(16.0) = 142.1 g/mol.

The number of grams required = (0.44 mol)(142.1 g/mol) = 63 g.

11. What [CrO42–] must be present in 0.00105 M AgNO3(aq) to just cause Ag2CrO4(s) to precipitate?

Answer

AgNO3(s) → Ag+(aq) + NO3–(aq)

Ag2CrO4(s) → ← 2 Ag+(aq) + CrO42–(aq)

Ksp = [Ag+]e2[CrO42–]e = 1.1×10–12


Initial0.00105 0

Change+0 +x

Equilibrium0.00105 x

(Note: the silver ion only arises from the silver nitrate; none comes from silver chromate because no silver chromate was introduced by the experiment.)

1.1×10–12 = [0.00105]2[x]

x = 1.0×10–6 M = the concentration of chromate required to just start precipitation.

12. In hard water, [Ca2+] is about 2.0×10–3 M. Water is fluoridated with 1.0 g F– per 1.0×103 L of water. Will CaF2(s) precipitate from hard water upon fluoridation?

Answer

Find the reaction quotient for CaF2 and compare it to Ksp.

CaF2(s) → ← Ca2+(aq) + 2 F–(aq)

Q = [Ca2+]init[F–]2init

[Ca2+]init = 2.0×10–3 M

[F–]init = (1.0 g)/(19.0 g/mol)/( 1.0×103 L) = 5.3×10–5 M

so Q = [2.0×10–3][ 5.3×10–5]2 = 5.6×10–12

Ksp = 5.3×10–9

Since Ksp > Q, there are not enough ions to satisfy the equilibrium condition so no precipitation occurs.

13. Calculate the solubility of Mg(OH)2 in a buffer solution that is 0.75 M NH3 and 0.50 M NH4Cl.

Answer

The solubility of magnesium hydroxide will depend upon the concentration of hydroxide present (hydroxide is a common ion between the buffer and the salt). Thus, the first step in the problem is to find the concentration of hydroxide in the buffer. Then, a typical solubility equilibrium problem can be solved. Assume 25 °C.

NH3(aq) + H2O(l) → ← OH–(aq) + NH4+(aq)

Kb = [NH4+]e[OH–]e                       [NH3]e = 1.8×10–5

Initial 0.75 0 0.50

Change – 0 +x +0

Equilibrium 0.75 x 0.50

The ammonia and ammonium ion concentrations do not change significantly because this is a buffer.

1.8×10–5 = (0.50)x              (0.75)

x = [OH–] = 2.7×10–5 M

Mg(OH)2(s) → ← Mg2+(aq) + 2 OH–(aq)

Ksp = [Mg2+]e[OH–]e2 = 1.8×10–11


Initial0 2.7×10–5

Change+x +0

Equilibriumx 2.7×10–5

In the buffer, the hydroxide ion concentration will be maintained constant.

1.8×10–11 = x(2.7×10–5)2

x = molar solubility of Mg(OH)2 (in the buffer) = 0.025 M

14. Which of the following complex ions would you expect to have the lowest [Ag+] in a solution that is 0.10 M in the complex ion and 1.0 M in the free ligand? Explain.

(a) [Ag(NH3)2]+

(b) [Ag(CN)2]–

(c) [Ag(S2O3)2]3–

Answer

Let L = ligand in each case, so the formation reaction is

Ag+(aq) + 2 Lm–(aq) → ← [AgL2](m–1)–(aq)

The mass action expression is:

Kf = [[AgL2](m–1)–]e                           [Ag+]e[L]e2


[Ag+]e = [[AgL2](m–1)–]e                           Kf[L]e2

Now the silver ion concentration can be calculated for each case:

(a) Kf = 1.6×107 so [Ag+] = [0.10]/(1.6×107)[1.0]2 = 6.3×10–9 M

(b) Kf = 5.6×1018 so [Ag+] = [0.10]/(5.6×1018)[1.0]2 = 1.8×10–20 M

(c) Kf = 1.7×1013 so [Ag+] = [0.10]/(1.7×1013)[1.0]2 = 5.9×10–15 M

The [Ag(CN)2]– solution has the lowest silver ion concentration.

Since all of these complex ions have the same stoichiometry, a comparison of the formation constants would have sufficed to answer the question: the largest formation constant leads to the lowest silver ion concentration.

15. When a few drops of concentrated Na2SO4(aq) are added to a dilute solution of AgNO3(aq), a white precipitate forms. When a small quantity of concentrated NH3(aq) is added to the mixture, the precipitate redissolves, resulting in a colorless solution. When this solution is made acidic with HNO3(aq), a white precipitate again appears. Write equations to represent these three observations.

Answer

Observation 1:

Na2SO4(aq) + 2 AgNO3(aq) → Ag2SO4(s) + 2 Na+(aq) + 2 NO3–(aq)

The white precipitate is silver sulfate.

Observation 2:

Ag2SO4(s) + 4 NH3(aq) → ← 2 [Ag(NH3)2]+(aq) + SO42–(aq)

The complex ion formation causes the silver sulfate to dissolve.

Observation 3:

2 [Ag(NH3)2]+(aq) + SO42–(aq) + 4 HNO3(aq) → Ag2SO4(s) + 4 NH4+(aq) + 4 NO3–(aq)

The acid/base reaction between nitric acid and ammonia causes dissociation of the complex ion, leaving the silver sulfate to reprecipitate.

16. What is the [Zn2+] in a solution that is 0.25 M in [Zn(NH3)4]2+ and has [NH3] equal to 1.50 M?

Answer

Zn2+(aq) + 4 NH3(aq) → ← [Zn(NH3)4]2+(aq)


Kf = [[Zn(NH3)4]2+]e                           [Zn2+]e[NH3]e4 = 4.1×108

Initial y z 0

Change – x –4x +x

Equilibrium y – x z – 4x x

The equilibrium conditions are given:

[[Zn(NH3)4]2+]e = 0.25 M = y – x

[NH3]e = 1.50 M = z – 4x

Fortunately, y and z do not need to be evaluated since the equilibrium conditions have been given.

4.1×108 = (0.25)                (x)(1.50)4


x = (0.25)                        (4.1×108)(1.50)4


x = [Zn2+] = 1.2×10–10 M