Dissolution of Lead Chloride Precipitate in Excess Hydrochloric Acid

To determine the cause of the dissolution of the precipitate, let us analyze the chemical reactions taking place step by step:

1. Precipitation Step: When aqueous solutions of lead(II) nitrate, $\text{Pb(NO}_3)_2$, and sodium chloride, $\text{NaCl}$, are mixed, a double displacement reaction occurs, forming a white precipitate of lead(II) chloride ($\text{PbCl}_2$): $\text{Pb(NO}_3)_2\text{(aq)} + 2\text{NaCl(aq)} \rightarrow \text{PbCl}_2\text{(s)}\downarrow + \,2\text{NaNO}_3\text{(aq)}$

2. Dissolution Step in Excess Hydrochloric Acid ($\text{HCl}$): When concentrated hydrochloric acid ($\text{HCl}$) is added to the mixture, the high concentration of chloride ions ($\text{Cl}^-$) drives a complexation reaction. Lead(II) chloride reacts with additional chloride ions to form a soluble coordination complex ion, tetrachloroplumbate(II): $\text{PbCl}_2\text{(s)} + 2\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{PbCl}_4]^{2-}\text{(aq)}$

Because the complex ion $[\text{PbCl}_4]^{2-}$ is soluble in water, the solid precipitate of $\text{PbCl}_2$ dissolves upon the addition of excess $\text{HCl}$.

Therefore, the dissolution of the precipitate is due to the formation of $[\text{PbCl}_4]^{2-}$.

Correct Answer: (C)