An inorganic salt solution gives a white precipitate when treated...

The formation of a white precipitate upon the addition of $(\text{NH}_4)_2\text{CO}_3$ in the presence of $\text{NH}_4\text{Cl}$ and $\text{NH}_4\text{OH}$ indicates the presence of a Group V cation ($\text{Ba}^{2+}$, $\text{Sr}^{2+}$, or $\text{Ca}^{2+}$).

The white precipitate is the metal carbonate ($\text{MCO}_3$), which dissolves in dilute acetic acid to form the corresponding soluble metal acetate.

Among the Group V cations, only $\text{Ba}^{2+}$ forms a yellow precipitate of barium chromate ($\text{BaCrO}_4$) upon the addition of aqueous $\text{K}_2\text{CrO}_4$.

$\text{Sr}^{2+}$ and $\text{Ca}^{2+}$ do not precipitate with chromate under these conditions. Although $\text{Pb}^{2+}$ also forms a yellow chromate precipitate, it belongs to Group I/II and its initial precipitation behavior does not match the Group V scheme. Thus, the cation is $\text{Ba}^{2+}$.

Answer: $\text{Ba}^{2+}$