Answer
$CaC{O_3}$ is more soluble is acidic solution than in pure water. That happens because the $C{O_3}^{2-}$ 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:
- $Ca^{2+}$: No significant acidity/basicity;
- $C{O_3}^{2-}$: Weak base. (It is the conjugate base of $HC{O_3}^-$, which is a weak acid);
2. Since we have a base in the compound, if we put it on acidic solution, this reaction will happen:
$C{O_3}^{2-}(aq) + H_3O^+(aq) -- \gt HC{O_3}^-(aq) + H_2O(l)$
Which means that, the concentration of $C{O_3}^{2+}$ will decrease, moving the equilibrium of:
$CaC{O_3}(s) \lt -- \gt Ca^{2+}(aq) + C{O_3}^{2-}(aq)$
To the right, consuming more $Ca{CO_3}(s)$, making the compound more soluble.
