JEE Advance - Chemistry (2022 - Paper 2 Online - No. 17)
The reaction of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{NaCl}$ in water produces a precipitate that dissolves upon the addition of $\mathrm{HCl}$ of appropriate concentration. The dissolution of the precipitate is due to the formation of
$\mathrm{PbCl}_{2}$
$\mathrm{PbCl}_{4}$
$\left[\mathrm{PbCl}_{4}\right]^{2-}$
$\left[\mathrm{PbCl}_{6}\right]^{2-}$
Explanation
$\mathrm{Pb}^{2+}$ on reaction with $\mathrm{Cl}^{-}$, produces white precipitate of $\mathrm{PbCl}_2$
$$ \mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{PbCl}_2 \downarrow $$ (White)
This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion
$$ \mathrm{PbCl}_2 \downarrow+2 \mathrm{Cl}^{-} \longrightarrow\left[\mathrm{PbCl}_4\right]^{2-} $$
$$ \mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{PbCl}_2 \downarrow $$ (White)
This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion
$$ \mathrm{PbCl}_2 \downarrow+2 \mathrm{Cl}^{-} \longrightarrow\left[\mathrm{PbCl}_4\right]^{2-} $$
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