Question

$PbCl_2$ is insoluble in cold water. Addition of $HCl$ increases its solubility due

• A. Formation of soluble complex anions like $[PbCl_3]^-$
• B. Oxidation of $Pb(II)$ to $Pb(IV)$
• C. Formation of $[Pb(H_2O)_6]^{2+}$
• D. Formation of polymeric lead complexes

Answer 1

Answer: (A)

$PbCl _{2}$ react with $HCl$ as follows
$P b C l_{2}(s)+C l^{-} \xrightarrow{\text { Cold }}\left[P b C l_{3}\right]^{-}(a q) $
$P b C l_{2}(s)+2 C l^{-} \xrightarrow[\text{of HCl}]{\text { Excess }}\left[P b C l_{4}\right]^{2-}(a q)$
Thus, addition of excess of $C l^{-}$ions change the $P b C l_{2}$ as soluble complex of $\left[ PbCl _{4}\right]^{-2}$.
Hence, becomes soluble.