Question

When HCl is passed through a saturated solution of common salt, pure NaCl is precipitated because:

• A. HCl is highly soluble in water
• B. the ionic product $\left[N{a}^{\oplus }\right]\left[C{l}^{⊝}\right]$ exceeds its solubility product $\left(K_{sp}\right)$.
• C. the $K_{sp}$ of NaCI is lowered by the presence of $C{l}^{⊝}$ ions
• D. HCl causes precipitation

Answer 1

Answer: (A), (B), (D)

HCl is highly soluble in water
HCl causes precipitation
the ionic product $\left[N{a}^{\oplus }\right]\left[C{l}^{⊝}\right]$ exceeds its solubility product $\left(K_{sp}\right)$.

When you have a saturated NaCl solution you have the equilibrium:

$NaCl\left(s\right)\leftrightharpoons N{a}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

This equilibrium obeys LeChatelier's principle,

When you add more $C{l}^{-}$ to the already saturated solution the equilibrium must shift to the left forming more NaCl(s).

When your solution contains no solid phase and all of the NaCl is present as ions in solution then you are able to produce very pure NaCl(s) by bubbling HCl(g) into the saturated solution of NaCl.