Question

The percentage hydrolysis of \( 0.15\,\text{M} \) solution of ammonium acetate.
Given: \( K_a \) for \( \text{CH}_3\text{COOH} = 1.8 \times 10^{-5} \) and \( K_b \) for \( \text{NH}_3 = 1.8 \times 10^{-5} \).

• A. 0.556
• B. 4.72
• C. 9.38
• D. 5.56

Hint:

Hydrolysis of salt of weak acid and weak base depends on Kₐ, Kb and concentration.

Answer 1

The Correct Option is C

Step 1: For salt of weak acid and weak base:
\( K_h = \dfrac{K_w}{K_a K_b} \) 

Step 2: Degree of hydrolysis:
\( h = \sqrt{\dfrac{K_h}{C}} \) 

Step 3: Substituting values gives:
\( h \approx 0.0938 \Rightarrow 9.38\% \)