The reaction of Pb(NO_3)_2 and NaCl in water | vedclass

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Question

(C) The reaction between $Pb(NO_3)_2$ and $NaCl$ produces a white precipitate of $PbCl_2$ according to the equation:$Pb(NO_3)_2(aq) + 2NaCl(aq) \right

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$[PbCl_4]^{2-}$

Vedclass · Answer

(C) The reaction between $Pb(NO_3)_2$ and $NaCl$ produces a white precipitate of $PbCl_2$ according to the equation:$Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow PbCl_2(s) + 2NaNO_3(aq)$When excess $HCl$ is added to this precipitate,it dissolves due to the formation of a soluble complex ion,tetrachloroplumbate$(II)$,represented as $[PbCl_4]^{2-}$:$PbCl_2(s) + 2Cl^-(aq) \rightarrow [PbCl_4]^{2-}(aq)$Thus,the dissolution is due to the formation of $[PbCl_4]^{2-}$.

The reaction of $Pb(NO_3)_2$ and $NaCl$ in water produces a precipitate that dissolves upon the addition of $HCl$ of appropriate concentration. The dissolution of the precipitate is due to the formation of

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