The problems are on the solubility product Ksp worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The solubility product Ksp is the equilibrium constant for a sparingly soluble ionic solid dissolving, such as AgCl(s) ⇌ Ag⁺ + Cl⁻, with Ksp = [Ag⁺][Cl⁻]. These worked answers correct the 20 problems of the AP Chemistry worksheet on solubility product Ksp, from write the Ksp expressions to error in the setup: each solution shows the formula, the substitution and the final answer with its unit.
Each answer starts with the dissolving equation, then the Ksp expression, then the substitution. Final values are in bold. Check that your answers use the right power of s, because that is the most common source of lost marks.
Level: AP Chemistry · Curriculum: AP Chem Unit 7
Problem 1 Write the Ksp expressions
(a) BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻; Ksp = [Ba²⁺][SO₄²⁻].(b) PbI₂(s) ⇌ Pb²⁺ + 2I⁻; Ksp = [Pb²⁺][I⁻]².
(c) Ag₃PO₄(s) ⇌ 3Ag⁺ + PO₄³⁻; Ksp = [Ag⁺]³[PO₄³⁻].
Problem 2 Solubility of silver bromide
s = √(5.4 × 10⁻¹³) = 7.35 × 10⁻⁷ mol/dm³.Problem 3 Precipitate or not
Q = 3.0 × 10⁻⁹ is larger than Ksp = 1.8 × 10⁻¹⁰, so the solution is supersaturated and a precipitate of AgCl forms until Q falls to Ksp.Problem 4 Ksp from mass solubility
1. s = 0.043 / 303.3 = 1.42 × 10⁻⁴ mol/dm³.2. Ksp = s² = 2.01 × 10⁻⁸.
Problem 5 Ksp of calcium carbonate
1. Ksp = s² = (7.1 × 10⁻⁵)² = 5.04 × 10⁻⁹.2. m = 7.1 × 10⁻⁵ × 100.1 = 7.1 mg/dm³.
Problem 6 Which salts like acid?
Acid helps when the anion is basic. CO₃²⁻ and OH⁻ react with H₃O⁺; Cl⁻ and SO₄²⁻ are very weak bases. CaCO₃ and Mg(OH)₂ dissolve better.Problem 7 Lead(II) iodide
1. Ksp = 4s³, so s = ³√(2.45 × 10⁻⁹) = 1.348 × 10⁻³ mol/dm³.2. [I⁻] = 2s = 2.696 × 10⁻³ mol/dm³.
Problem 8 Silver chromate
s = ³√(1.1 × 10⁻¹² / 4) = ³√(2.75 × 10⁻¹³) = 6.50 × 10⁻⁵ mol/dm³. [Ag⁺] = 2s = 1.30 × 10⁻⁴ mol/dm³.Problem 9 Ksp of calcium hydroxide
[Ca²⁺] = 0.011 and [OH⁻] = 0.022 mol/dm³. Ksp = 0.011 × 0.022² = 5.32 × 10⁻⁶. pOH = 1.66, so pH = 12.34.Problem 10 Common ion effect
[Cl⁻] ≈ 0.050, so s = 1.8 × 10⁻¹⁰ / 0.050 = 3.60 × 10⁻⁹ mol/dm³. In pure water s = 1.34 × 10⁻⁵ mol/dm³, so the solubility falls by a factor of 3727.Problem 11 CaF₂ in fluoride solution
[F⁻] ≈ 0.020, so s = [Ca²⁺] = Ksp/[F⁻]² = 3.5 × 10⁻¹¹ / (0.020)² = 8.75 × 10⁻⁸ mol/dm³. In pure water s = 2.06 × 10⁻⁴ mol/dm³, about 2355 times larger.Problem 12 Mixing two solutions
1. The volume doubles to 200.0 cm³, so [Ag⁺] = 1.0 × 10⁻⁴ and [Cl⁻] = 2.0 × 10⁻⁴ mol/dm³.2. Q = (1.0 × 10⁻⁴)(2.0 × 10⁻⁴) = 2.00 × 10⁻⁸.
3. Q is greater than Ksp, so AgCl precipitates.
Problem 13 Magnesium hydroxide and acid
Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻. Added H₃O⁺ reacts with OH⁻: H₃O⁺ + OH⁻ → 2H₂O. Removing OH⁻ makes Q smaller than Ksp, so the equilibrium shifts right and the solid dissolves.Problem 14 Which precipitates first?
1. [Ag⁺] = 1.8 × 10⁻¹⁰ / 0.010 = 1.80 × 10⁻⁸ mol/dm³.2. [Ag⁺] = √(1.1 × 10⁻¹² / 0.010) = 1.05 × 10⁻⁵ mol/dm³.
3. AgCl needs far less Ag⁺, so AgCl precipitates first.
Problem 15 Barium sulfate in a drink
s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ mol/dm³. n = 1.05 × 10⁻⁵ × 0.500 = 5.24 × 10⁻⁶ mol. m = 5.24 × 10⁻⁶ × 233.4 = 1.22 mg.Problem 16 Buffered magnesium hydroxide
1. s = ³√(5.6 × 10⁻¹² / 4) = 1.12 × 10⁻⁴ mol/dm³; [OH⁻] = 2.24 × 10⁻⁴; pH = 10.35.2. [OH⁻] = 1.0 × 10⁻⁴, so s = 5.6 × 10⁻¹² / (1.0 × 10⁻⁴)² = 5.60 × 10⁻⁴ mol/dm³.
3. The buffer holds [OH⁻] below the value the dissolving would produce (2.2 × 10⁻⁴), so more solid dissolves to reach Ksp.
Problem 17 Separating by precipitation
1. [Ag⁺] = 1.05 × 10⁻⁵ mol/dm³.2. [Cl⁻] = 1.8 × 10⁻¹⁰ / 1.05 × 10⁻⁵ = 1.72 × 10⁻⁵ mol/dm³.
3. Remaining = 0.17 % of the original 0.010 mol/dm³. So about 99.8 % of the chloride has been removed first, which is a good separation.
Problem 18 Calcium fluoride precipitation
1. The volume doubles, so [Ca²⁺] = 0.0020 and [F⁻] = 5.0 × 10⁻⁴ mol/dm³.2. Q = (0.0020)(5.0 × 10⁻⁴)² = 5.00 × 10⁻¹⁰, which is greater than Ksp, so CaF₂ precipitates.
3. [F⁻] = √(3.5 × 10⁻¹¹ / 0.0020) = 1.32 × 10⁻⁴ mol/dm³.
Problem 19 Iron(III) hydroxide and pH
1. Ksp = [Fe³⁺][OH⁻]³.2. [OH⁻] = 1.0 × 10⁻⁷, so [Fe³⁺] = 2.8 × 10⁻³⁹ / (1.0 × 10⁻⁷)³ = 2.8 × 10⁻¹⁸ mol/dm³.
3. [OH⁻] = 1.0 × 10⁻¹¹, so [Fe³⁺] = 2.8 × 10⁻⁶ mol/dm³.
4. The solubility depends on [OH⁻]³. At pH 7 almost no iron stays dissolved, so brown Fe(OH)₃ precipitates. At pH 3 it is a trillion times more soluble.
Problem 20 Error in the setup
1. The student used the 1:1 formula s = √Ksp, but Ag₂CrO₄ gives 2Ag⁺ + CrO₄²⁻, so Ksp = (2s)²s = 4s³.2. s = ³√(2.75 × 10⁻¹³) = 6.50 × 10⁻⁵ mol/dm³.
3. Factor = 6.503 × 10⁻⁵ / 1.049 × 10⁻⁶ = about 62 times too small.
