Answer
$Mg(OH)_2$ is more soluble is acidic solution than in pure water. That happens because the $OH^-$ 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:
- $Mg^{2+}$: No significant acidity/basicity;
- $OH^{-}$: Strong base.
2. Since we have a base in the compound, if we put it on acidic solution, this reaction will happen:
$OH^{-}(aq) + H_3O^+(aq) -- \gt 2H_2O(l)$
Which means that, the concentration of $OH^-$ will decrease, moving the equilibrium of:
$Mg(OH)_2(s) \lt -- \gt Mg^{2+}(aq) + OH^{-}(aq)$
To the right, consuming more $Mg(OH)_2(s)$, making the compound more soluble.
