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

The Correct Option is A

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