2017 H2 Chem P1 Q15 — Solubility Solution

The answer is A (\(1.7 \times 10^{-14}\)).
\(\text{PbCrO}_4(\text{s}) \rightleftharpoons \text{Pb}^{2+}(\text{aq}) + \text{CrO}_4^{2-}(\text{aq})\)
Each mole that dissolves gives 1 mol of each ion, so \([\text{Pb}^{2+}] = [\text{CrO}_4^{2-}] = 1.3 \times 10^{-7} \text{ mol dm}^{-3}\).
\(K_\text{sp} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}] = (1.3 \times 10^{-7})^2 = 1.69 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}\), which is \(1.7 \times 10^{-14}\).
• B is wrong: \(6.8 \times 10^{-14} = (2.6 \times 10^{-7})^2\), which takes each ion concentration as twice the solubility. \(\text{PbCrO}_4\) gives only one of each ion per formula unit.
• C is wrong: \(1.3 \times 10^{-7}\) is the solubility itself, not the product of the two ion concentrations.
• D is wrong: \(2.6 \times 10^{-7}\) is twice the solubility (the sum of the two ion concentrations), not their product.