Concept involved: common ion effect and solubility in presence of concentrated HCl

Passing $\mathrm{HCl}_{(g)}$ into an aqueous chloride solution does two things:

• It increases the concentration of $\mathrm{Cl^-}$ in solution.
• It also changes the solvent environment because dissolved $\mathrm{HCl}$ makes the medium strongly acidic and less favorable for hydration of some salts.

Whether a chloride precipitates depends on how much its solubility decreases.

Examine Statement II first: saturated $\mathrm{NaCl}$ solution + $\mathrm{HCl}_{(g)}$

Dissolution equilibrium: \mathrm{NaCl}_{(s)} \rightleftharpoons Na^+_{(aq)} + Cl^-_{(aq)}}

On passing $\mathrm{HCl}$ gas, the concentration of $\mathrm{Cl^-}$ increases, so by common ion effect the equilibrium shifts left.

Hence $\mathrm{NaCl}$ precipitates out from its saturated solution.

Therefore, Statement II is correct.

Now examine Statement I: saturated $\mathrm{BaCl_2}$ solution + $\mathrm{HCl}_{(g)}$

Dissolution equilibrium: \mathrm{BaCl_2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2Cl^-_{(aq)}}

If only common ion effect were important, increasing $\mathrm{Cl^-}$ could reduce solubility. However, $\mathrm{BaCl_2}$ is highly soluble in water, and under ordinary conditions passing $\mathrm{HCl}$ gas through its saturated solution does not produce noticeable white turbidity as in the case of salts whose solubility is significantly lowered.

Thus the statement claiming appearance of white turbidity is incorrect.

Evaluate options

• A: Both correct — false
• B: Statement I incorrect, Statement II correct — true
• C: Both incorrect — false
• D: Statement I correct, Statement II incorrect — false

Derived answer

The correct option is: $\boxed{\text{B}}$

Comparison with stored answer

Stored correct answer is D, but from equilibrium and common ion effect, Statement II is definitely correct, while Statement I is not.

So I disagree with the stored answer.