Answer
$CuS$ is more soluble is acidic solution than in pure water. That happens because the $S^{2-}$ is a base, and it reacts with the hydronium ions.
Work Step by Step
1. Identify the acidity/basicity of each ion in the compound:
- $Cu^{2+}$: Weak (Lewis) acid.
- $S^{2-}$: Strong base;
2. Since we have a strong base in the compound, if we put it on acidic solution, this reaction will happen:
$S^{2-}(aq) + H_3O^+(aq) -- \gt HS^-(aq) + H_2O(l)$
Which means that, the concentration of $S^{2+}$ will decrease, moving the equilibrium of:
$CuS(s) \lt -- \gt Cu^{2+}(aq) + S^{2-}(aq)$
To the right, consuming more $CuS(s)$, making the compound more soluble.
