2022 H2 Chem P1 Q16 — Solubility Solution

The answer is C.

\(\text{Ag}_2\text{SO}_4\text{(s)} \rightleftharpoons 2\text{Ag}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}\)

In water: \([\text{Ag}^+] = 0.032\) mol dm\(^{-3}\), so \([\text{SO}_4^{2-}] = \frac{1}{2} \times 0.032 = 0.016\) mol dm\(^{-3}\).
\(K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}] = (0.032)^2 \times 0.016 = 1.6384 \times 10^{-5}\) mol\(^3\) dm\(^{-9}\)

In 0.50 mol dm\(^{-3}\) \(\text{Na}_2\text{SO}_4\): let the solubility of \(\text{Ag}_2\text{SO}_4\) be s mol dm\(^{-3}\). Then \([\text{Ag}^+] = 2s\) and \([\text{SO}_4^{2-}] = 0.50 + s\).
\(K_{sp} = (2s)^2(0.50 + s) = 1.6384 \times 10^{-5}\)
The common \(\text{SO}_4^{2-}\) ion suppresses the solubility, so \(s \ll 0.50\) and \(0.50 + s \approx 0.50\):
\((2s)^2(0.50) = 1.6384 \times 10^{-5}\)
\(s = \sqrt{\frac{1.6384 \times 10^{-5}}{4 \times 0.50}} = 2.86 \times 10^{-3}\) mol dm\(^{-3}\) (small compared with 0.50, so the approximation holds)

\([\text{Ag}^+] = 2s = 5.72 \times 10^{-3} \approx 5.7 \times 10^{-3}\) mol dm\(^{-3}\)

• A is wrong: \(3.3 \times 10^{-5}\) is \(\frac{1.6384 \times 10^{-5}}{0.50}\), which is \([\text{Ag}^+]^2\); the square root has not been taken.
• B is wrong: \(6.6 \times 10^{-5}\) is twice the value in A, again without taking the square root.
• D is wrong: \(8.1 \times 10^{-3}\) comes from taking \([\text{SO}_4^{2-}]\) in water as 0.032 instead of 0.016 mol dm\(^{-3}\), giving \(K_{sp} = (0.032)^3 = 3.28 \times 10^{-5}\) and \([\text{Ag}^+] = \sqrt{\frac{3.28 \times 10^{-5}}{0.50}} = 8.1 \times 10^{-3}\) mol dm\(^{-3}\).