Given below are two statements: Statement I: On passing HCl gas through a saturated solution of BaCl2, at room…

⚠️ Note: This question has conflicting answers across all online and offline sources. We provide an unusually detailed explanation below so you can understand what is actually happening chemically.

Step 1 — Evaluate Statement I: Passing HCl gas through saturated $\ce{BaCl2}$ causes white turbidity

At first glance, since $\ce{BaCl2}$ is "soluble," many sources say no precipitation occurs. However, the reality is more nuanced:

The $K_{sp}$ of $\ce{BaCl2}$ is not a tiny number. Unlike sparingly soluble salts (e.g., $\ce{AgCl}$, $K_{sp} = 1.8 \times 10^{-10}$), $\ce{BaCl2}$ is highly soluble (~370 g/kg water at 25°C), giving a thermodynamic $K_{sp}^\circ \approx 101.1$ (using activities with mean activity coefficient $\gamma_{\pm} \approx 1.65$).

Numerical proof:

At saturation in 1 kg water: $[\ce{Ba^{2+}}] = 1.779$ m, $[\ce{Cl^-}] = 3.558$ m

Ionic product $Q = (1.779 \times 1.65) \times (3.558 \times 1.65)^2 \approx 101.1 = K_{sp}$ (system at equilibrium) ✓

Now bubble HCl gas until 1 mol $\ce{Cl^-}$ is added:

New $[\ce{Cl^-}] = 3.558 + 1.0 = 4.558$ m

New $Q = (1.779 \times 1.65) \times (4.558 \times 1.65)^2 \approx 166.0 > K_{sp}$

Since $Q > K_{sp}$, the solution becomes supersaturated → $\ce{Ba^{2+}}$ and $\ce{Cl^-}$ ions leave solution → white precipitate of $\ce{BaCl2 \cdot 2H2O}$ crystals forms ✓

Additionally, HCl gas is extremely hygroscopic — it "steals" water molecules for its own hydration, reducing the solvent available for $\ce{Ba^{2+}}$ and $\ce{Cl^-}$ ions (salting-out effect), further driving precipitation.

Statement I is TRUE ✓

Step 2 — Evaluate Statement II: Passing HCl gas through saturated NaCl causes precipitation (common ion effect)

$\ce{NaCl}$ has a very high $K_{sp}$ (~36–38 at 25°C). Adding HCl gas:

• Increases $[\ce{Cl^-}]$ dramatically
• $Q_{sp} = [\ce{Na^+}][\ce{Cl^-}] > K_{sp}$ → $\ce{NaCl}$ precipitates

This is the classic common ion effect and is the standard industrial method for purifying common salt (impurities like $\ce{MgCl2}$ and $\ce{CaCl2}$ remain dissolved as their ionic products don't reach their $K_{sp}$ limits as quickly).

Statement II is TRUE ✓

Step 3 — Conclusion

Both Statement I and Statement II are correct.

The key insight for Statement I: $\ce{BaCl2}$ precipitation with HCl occurs because even though $\ce{BaCl2}$ is "highly soluble," its $K_{sp}$ is a finite number (~101). Adding excess $\ce{Cl^-}$ via HCl gas pushes $Q > K_{sp}$, causing precipitation. The mechanism involves both the common ion effect and hygroscopic dehydration by HCl.

Answer: Option (a) — Both Statement I and Statement II are correct

Key Points to Remember:

• "Highly soluble" does NOT mean $K_{sp} = \infty$; $\ce{BaCl2}$ has $K_{sp}^\circ \approx 101$
• Adding HCl gas to saturated $\ce{BaCl2}$: $Q$ exceeds $K_{sp}$ → white precipitate ($\ce{BaCl2 \cdot 2H2O}$)
• HCl is hygroscopic → salting-out effect further drives precipitation
• NaCl + HCl gas: classic common ion effect → NaCl precipitates (used in industrial salt purification)
• This question is genuinely controversial in JEE/NEET literature; the thermodynamically correct answer is that both statements are true