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Difficulty: HardSolubility, Solubility Curves, and Solubility Product (Ksp)

The solubility product (KspK_{sp}) of lead(II) chloride (PbCl2\text{PbCl}_2) at 25∘C25^\circ\text{C} is 3.2×10−5 mol3 dm−93.2 \times 10^{-5}\text{ mol}^3\text{ dm}^{-9}. What is the concentration of chloride ions (Cl−\text{Cl}^-) in mol dm−3\text{mol dm}^{-3} in a saturated solution of lead(II) chloride at 25∘C25^\circ\text{C}?

  1. A
    2.0×10−2 mol dm−32.0 \times 10^{-2}\text{ mol dm}^{-3}
  2. 4.0×10−2 mol dm−34.0 \times 10^{-2}\text{ mol dm}^{-3}Answer
  3. C
    1.42 mol dm−31.42\text{ mol dm}^{-3}
  4. D
    3.17×10−2 mol dm−33.17 \times 10^{-2}\text{ mol dm}^{-3}

Answer

The concentration of chloride ions in the saturated solution is 4.0×10−2 mol dm−34.0 \times 10^{-2}\text{ mol dm}^{-3}.
For the dissolution equilibrium PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq), the solubility product expression is Ksp=[Pb2+][Cl−]2K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2. Letting ss equal the molar solubility of PbCl2\text{PbCl}_2, [Pb2+]=s[\text{Pb}^{2+}] = s and [Cl−]=2s[\text{Cl}^-] = 2s. Substituting into KspK_{sp} gives Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3. Given Ksp=3.2×10−5K_{sp} = 3.2 \times 10^{-5}, solving 4s3=3.2×10−54s^3 = 3.2 \times 10^{-5} yields s3=8.0×10−6s^3 = 8.0 \times 10^{-6} and s=2.0×10−2 mol dm−3s = 2.0 \times 10^{-2}\text{ mol dm}^{-3}. The chloride ion concentration is [Cl−]=2s=4.0×10−2 mol dm−3[\text{Cl}^-] = 2s = 4.0 \times 10^{-2}\text{ mol dm}^{-3}.

Step-by-Step Solution

1
Write the solubility equilibrium equation and express KspK_{sp} in terms of molar solubility (ss).
PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq), so Ksp=[Pb2+][Cl−]2=s(2s)2=4s3K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s(2s)^2 = 4s^3.
Dissolution of one mole of PbCl2\text{PbCl}_2 produces one mole of Pb2+\text{Pb}^{2+} ions and two moles of Cl−\text{Cl}^- ions.
2
Substitute the given KspK_{sp} value and calculate the molar solubility (ss).
3.2×10−5=4s3  ⟹  s3=8.0×10−6  ⟹  s=2.0×10−2 mol dm−33.2 \times 10^{-5} = 4s^3 \implies s^3 = 8.0 \times 10^{-6} \implies s = 2.0 \times 10^{-2}\text{ mol dm}^{-3}.
Dividing KspK_{sp} by 4 gives s3s^3, and taking the cube root yields ss.
3
Determine the concentration of chloride ions ([Cl−][\text{Cl}^-]).
[Cl−]=2s=2×(2.0×10−2 mol dm−3)=4.0×10−2 mol dm−3[\text{Cl}^-] = 2s = 2 \times (2.0 \times 10^{-2}\text{ mol dm}^{-3}) = 4.0 \times 10^{-2}\text{ mol dm}^{-3}.
Since two moles of chloride ions are released per mole of salt dissolved, [Cl−][\text{Cl}^-] equals 2s2s.

Key Concept

Solubility product (KspK_{sp}) calculation for MX2MX_2 type salts and stoichiometric determination of ion concentrations.
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Topics

Solubility, Solubility Curves, and Solubility Product (Ksp)

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