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Equilibria: Question 5

Syllabus 25.2

Structured A2 7 marks

Lead(II) iodide, PbI2\text{PbI}_2, is a sparingly soluble ionic solid that establishes the following equilibrium in contact with water:

PbI2(s)⇌Pb2+(aq)+2I−(aq)\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)

(a) Write the expression for the solubility product, KspK_{sp}, of PbI2\text{PbI}_2. [1]

(b) The molar solubility of PbI2\text{PbI}_2 in pure water at 298 K is 1.5×10−3 mol dm−31.5\times10^{-3}\ \text{mol dm}^{-3}. Calculate KspK_{sp} for PbI2\text{PbI}_2, including its units. [3]

(c) Calculate the molar solubility of PbI2\text{PbI}_2 in a 0.10 mol dm−30.10\ \text{mol dm}^{-3} solution of potassium iodide, using your value of KspK_{sp} from (b). Comment briefly on how this compares with the solubility in pure water. [3]

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Worked solution

Part (a): Writing the Ksp expression

Ksp=[Pb2+(aq)][I−(aq)]2K_{sp}=[\text{Pb}^{2+}(aq)][\text{I}^-(aq)]^2

Part (b): Calculating Ksp from the solubility in water

Let the molar solubility be s=1.5×10−3 mol dm−3s=1.5\times10^{-3}\ \text{mol dm}^{-3}. Since each mole of PbI2\text{PbI}_2 that dissolves gives 1 mole of Pb2+\text{Pb}^{2+} and 2 moles of I−\text{I}^-: [Pb2+]=s=1.5×10−3 mol dm−3[\text{Pb}^{2+}]=s=1.5\times10^{-3}\ \text{mol dm}^{-3} [I−]=2s=3.0×10−3 mol dm−3[\text{I}^-]=2s=3.0\times10^{-3}\ \text{mol dm}^{-3}

Substituting into the expression from part (a): Ksp=(1.5×10−3)×(3.0×10−3)2=(1.5×10−3)×(9.0×10−6)K_{sp}=(1.5\times10^{-3})\times(3.0\times10^{-3})^2=(1.5\times10^{-3})\times(9.0\times10^{-6}) Ksp=1.35×10−8 mol3 dm−9K_{sp}=1.35\times10^{-8}\ \text{mol}^3\ \text{dm}^{-9}

(Check: (3.0×10−3)2=9.0×10−6(3.0\times10^{-3})^2=9.0\times10^{-6}, and 1.5×9.0=13.51.5\times9.0=13.5, giving 13.5×10−9=1.35×10−813.5\times10^{-9}=1.35\times10^{-8}, consistent. Units: mol dm−3×(mol dm−3)2=mol3 dm−9\text{mol dm}^{-3}\times(\text{mol dm}^{-3})^2=\text{mol}^3\ \text{dm}^{-9}.)

Part (c): Solubility in KI solution (common-ion effect)

In 0.10 mol dm−30.10\ \text{mol dm}^{-3} potassium iodide, KI is fully dissociated and provides far more I−\text{I}^- than the small amount released by dissolving PbI2\text{PbI}_2, so [I−]≈0.10 mol dm−3[\text{I}^-]\approx0.10\ \text{mol dm}^{-3}. Let the new molar solubility of PbI2\text{PbI}_2 be s′=[Pb2+]s'=[\text{Pb}^{2+}]:

Ksp=s′×(0.10)2=s′×0.010=1.35×10−8K_{sp}=s'\times(0.10)^2=s'\times0.010=1.35\times10^{-8} s′=1.35×10−80.010=1.35×10−6 mol dm−3s'=\frac{1.35\times10^{-8}}{0.010}=1.35\times10^{-6}\ \text{mol dm}^{-3}

(Check: 1.35×10−8÷1.0×10−2=1.35×10−61.35\times10^{-8}\div1.0\times10^{-2}=1.35\times10^{-6}. Consistent.)

Comparison: 1.35×10−6 mol dm−31.35\times10^{-6}\ \text{mol dm}^{-3} is roughly a thousand times smaller than the 1.5×10−3 mol dm−31.5\times10^{-3}\ \text{mol dm}^{-3} solubility in pure water (a factor of about 1.1×1031.1\times10^{3}). Adding the common ion I−\text{I}^- shifts the position of the dissolution equilibrium to the left (Le Chatelier’s principle), suppressing further dissolution of PbI2\text{PbI}_2. This is the common-ion effect.

Final answers

  • (a) Ksp=[Pb2+][I−]2K_{sp}=[\text{Pb}^{2+}][\text{I}^-]^2
  • (b) Ksp=1.35×10−8 mol3 dm−9K_{sp}=1.35\times10^{-8}\ \text{mol}^3\ \text{dm}^{-9}
  • (c) s′=1.35×10−6 mol dm−3s'=1.35\times10^{-6}\ \text{mol dm}^{-3}, roughly a thousand times less soluble than in pure water, due to the common-ion effect