Question

What will be the change in acidity if:
(i) CuSO₄ is added to saturated (NH₄)₂SO₄ solution,
(ii) SbF₅ is added to anhydrous HF?

• A. Increases, Increases
• B. Decreases, Decreases
• C. Increases, Decreases
• D. Decreases, Increases

Hint:

The effect of added compounds on acidity depends on whether the compound acts as a Lewis acid or base, or influences ion dissociation.

Answer 1

The Correct Option is A

Part (i): CuSO₄ added to saturated (NH₄)₂SO₄ solution

When CuSO₄ is added to a saturated solution of (NH₄)₂SO₄, it introduces Cu²⁺ ions into the solution. Cu²⁺ is a Lewis acid and reacts with water, increasing the concentration of H⁺ ions. This reaction results in an increase in acidity:

\( \text{Cu}^{2+} + \text{H}_2\text{O} \rightarrow [\text{Cu(H}_2\text{O)}_6]^{2+} \rightarrow [\text{Cu(H}_2\text{O)}_5(\text{OH})]^+\ + \text{H}^+ \)

Part (ii): SbF₅ added to anhydrous HF

SbF₅ is a strong Lewis acid and forms the superacid HF-SbF₅ when mixed with HF. This combination greatly increases the acidity of the solution as it stabilizes the fluoride ion (F⁻), shifting the equilibrium towards producing more H⁺ ions:

\( \text{HF} + \text{SbF}_5 \rightarrow \text{HSbF}_6\ + \text{H}^+ \)

In both cases, the acidity of the system increases.

Conclusion: The correct answer is: Increases, Increases.

Answer 2

Correct Answer:

Option 1: Increases, Increases

Explanation:

(i) CuSO4 added to saturated (NH4)2SO4 solution:

CuSO4 introduces Cu²⁺ ions. These ions react with NH3 produced from the equilibrium of NH4⁺ in solution, forming [Cu(NH3)4]²⁺. This removes NH3, shifting the NH4⁺ equilibrium to produce more H3O⁺, increasing acidity.

(ii) SbF5 added to anhydrous HF:

SbF5 acts as a Lewis acid, reacting with F^- ions from HF to form SbF6^-. This removes F^-, shifting the HF dissociation equilibrium to produce more H⁺, increasing acidity.