Why is pure NaCl precipitated when HCl gas is | vedclass

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(C) The dissolution of $NaCl$ is represented by the equilibrium: $NaCl_{(s)} \rightleftharpoons Na^{ }_{(aq)}   Cl^{-}_{(aq)}$.When $HCl$ gas is passe

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The value of $[Na^{+}] [Cl^{-}]$ becomes greater than $K_{sp}$ of $NaCl$

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(C) The dissolution of $NaCl$ is represented by the equilibrium: $NaCl_{(s)} \rightleftharpoons Na^{ }_{(aq)}   Cl^{-}_{(aq)}$.When $HCl$ gas is passed through the saturated solution,it dissociates as: $HCl_{(g)}  \rightarrow H^{ }_{(aq)}   Cl^{-}_{(aq)}$.This increases the concentration of $Cl^{-}$ ions in the solution (Common Ion Effect).According to Le Chatelier's principle,the increase in $[Cl^{-}]$ shifts the equilibrium to the left.Consequently,the ionic product $[Na^{ }] [Cl^{-}]$ exceeds the solubility product constant $(K_{sp})$ of $NaCl$,causing the precipitation of pure $NaCl$.

Why is pure $NaCl$ precipitated when $HCl$ gas is passed through a saturated solution of $NaCl$?

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