2018 H2 Chem P1 Q13 — Periodic table Solution

The answer is A (aluminium sulfate).
In water a metal ion is surrounded by water ligands. A small, highly charged ion polarises the O-H bonds of these water ligands, so they lose \(\text{H}^+\) (hydrolysis), for example:
\([\text{Al}(\text{H}_2\text{O})_6]^{3+} \rightleftharpoons [\text{Al}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}^+\)
The higher the charge density \(\left(\propto \dfrac{\text{charge}}{\text{ionic radius}}\right)\), the greater the extent of hydrolysis and the more acidic the solution. Data Booklet ionic radii: \(\text{Al}^{3+}\) 0.050 nm, \(\text{Mg}^{2+}\) 0.065 nm, \(\text{Cu}^{2+}\) 0.073 nm, \(\text{Na}^+\) 0.095 nm.
\(\text{Al}^{3+}\) has the highest charge (3+) and the smallest radius, so it has the highest charge density, hydrolyses to the greatest extent and gives the highest \([\text{H}^+]\).
• B is wrong: \(\text{Cu}^{2+}\) has a lower charge and a larger radius than \(\text{Al}^{3+}\), so its charge density is lower and it hydrolyses less than \(\text{Al}^{3+}\).
• C is wrong: \(\text{Mg}^{2+}\) also has a lower charge and a larger radius than \(\text{Al}^{3+}\), so it hydrolyses less than \(\text{Al}^{3+}\).
• D is wrong: \(\text{Na}^+\) has only a 1+ charge and the largest radius, so it does not hydrolyse to any significant extent and its solution is not acidic.