Verified Solution • 2025–26 Session moderate

QJEEM24Cha798: JEE Main 2024 Chemistry - Class 11 JEEM24Cha798 Solution: Given below are two statements : Statement I : On passing HCl ( …

Given below are two statements :

Statement I : On passing HCl(g) through a saturated solution of BaCl2, 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 :

Multiple Choice Options

  • A Both Statement I and Statement II are correct
  • B Statement I is incorrect but Statement II is correct
  • C Both Statement I and Statement II are incorrect
  • D Statement I is correct but Statement II is incorrect

Step-by-Step Verified Solution

To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.


Statement I:

On passing HCl(g)\mathrm{HCl}_{(g)} through a saturated solution of BaCl2\mathrm{BaCl}_2 at room temperature, white turbidity appears.

Analysis:

Dissolution of HCl Gas:

When HCl(g)\mathrm{HCl}_{(g)} is bubbled through water, it dissolves and dissociates completely:

HCl(aq)→H(aq)++Cl(aq)−\mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}

This increases the concentration of Cl−\mathrm{Cl}^- ions in the solution.

Effect on BaCl₂ Solubility:

The solubility equilibrium of BaCl2\mathrm{BaCl}_2 in water is:

BaCl2↔Ba(aq)2++2Cl(aq)−\mathrm{BaCl}_2 \leftrightarrow \mathrm{Ba}^{2+}_{(aq)} + 2\mathrm{Cl}^-_{(aq)}

Adding more Cl−\mathrm{Cl}^- shifts the equilibrium to the left (Le Chatelier's Principle), causing BaCl2\mathrm{BaCl}_2 to precipitate.

The precipitation of BaCl2\mathrm{BaCl}_2 manifests as a white turbidity.

Conclusion:

Statement I is correct.


Statement II:

When HCl\mathrm{HCl} gas is passed through a saturated solution of NaCl\mathrm{NaCl} , sodium chloride is precipitated due to common ion effect.

Analysis:

Dissolution of HCl Gas:

Similar to before, HCl\mathrm{HCl} increases Cl−\mathrm{Cl}^- ion concentration.

Effect on NaCl Solubility:

The solubility equilibrium of NaCl\mathrm{NaCl} is:

NaCl↔Na(aq)++Cl(aq)−\mathrm{NaCl} \leftrightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}

However, NaCl\mathrm{NaCl} is highly soluble in water, and its solubility is not significantly affected by the common ion effect from Cl−\mathrm{Cl}^- .

The solubility product ( KspK_{sp} ) of NaCl\mathrm{NaCl} is large, and the addition of Cl−\mathrm{Cl}^- ions does not cause NaCl\mathrm{NaCl} to precipitate under normal conditions.

No precipitation occurs; the solution remains clear.

Conclusion:

Statement II is incorrect.


Final Answer:

Statement I is correct but Statement II is incorrect.

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