2023 H2 Chem P1 Q15 — Solubility Solution

The answer is B (1 and 2 only).
• Statement 1, correct. Mixing sodium ethanoate and silver nitrate gives a white precipitate of \(\text{CH}_3\text{COOAg}\), so silver ethanoate is insoluble (sparingly soluble) in water.
• Statement 2, correct. Silver ethanoate sets up \(\text{CH}_3\text{COOAg(s)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{Ag}^+\text{(aq)}\). When \(\text{Br}^-\) is added, the ionic product of AgBr exceeds its very small \(K_\text{sp}\) even at this low \([\text{Ag}^+]\), so cream AgBr precipitates. Removing \(\text{Ag}^+\) shifts the equilibrium to the right and the white precipitate dissolves. In the second experiment, adding ethanoate to AgBr causes no change, because the \([\text{Ag}^+]\) above AgBr is too low for silver ethanoate to precipitate. Both observations show AgBr is less soluble than silver ethanoate.
• Statement 3, incorrect. The cream precipitate is AgBr formed by precipitation, with no change in oxidation state. If ethanoate oxidised bromide, \(\text{Br}_2\) would form, and the second student would also have seen a change on adding ethanoate, but there was none.
• A is wrong: it includes statement 3, and no redox reaction takes place.
• C is wrong: it includes statement 3 and leaves out statement 1, which is shown by the white precipitate.
• D is wrong: it leaves out statement 2, which is needed to explain why the white precipitate turns into a cream one.