Chemistry · Ionic Equilibrium

JEE Main 2024 — 5 April, Shift 2 — Question 65

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II.

Statement I : On passing $\mathrm{HCl}{(\mathrm{g})}$ through a saturated solution of $\mathrm{BaCl}_{2}$ , at room temperature white turbidity appears.

Statement II : When HCl gas is passed through a saturated solution of NaCl , sodium chloride is precipitated due to common ion effect.

In the light of the above statements, choose the most appropriate answer from the options given below:

Option A:
Statement I is correct but Statement II is incorrect
Correct
Option B:
Both Statement I and Statement II are incorrect
Option C:
Statement I is incorrect but Statement II is correct
Option D:
Both Statement I and Statement II are correct

Answer: A

Step-by-step solution

For Statement I: BaCl₂ is sparingly soluble. Passing HCl(g) increases Cl⁻ concentration via dissociation: $\text{HCl}(g) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq)$ . Common ion effect reduces solubility: $\text{BaCl}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{Cl}^-(aq)$ . Ionic product exceeds Ksp, causing precipitation of BaCl₂ as white turbidity.

For Statement II: NaCl is highly soluble in water. Common ion effect from added Cl⁻ is insufficient to exceed its very high solubility product. No precipitation occurs; NaCl remains dissolved.

Thus, Statement I is correct, Statement II is incorrect. Correct option: A.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Sparingly Soluble Salts, Solubility Product & Precipitation Conditions