Q14 / 155 · WBJEE 2017 · Chemistry

$\mathrm{PbCl}_{2}$ is insoluble in cold water. Addition of $\mathrm{HCl}$ increases its solubility due to

  • Correct answer
    A
    formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
  • B
    oxidation of Pb(II) to PB(iV)
  • C
    formation of $\left[\mathrm{Pb}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2}$
  • D
    formation of polymeric lead complexes

Solution

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

Hence, the correct answer is option: 'A'.

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