Question

The percentage hydrolysis of 0.15M solution of ammonium acetate, Kₐ for CH₃COOH=1.8×10⁻5 and Kb for NH₃=1.8×10⁻5 is:

• A. \(0.556\)
• B. \(4.72\)
• C. \(9.38\)
• D. 5.56

Hint:

For salts of weak acid and weak base: h=√((Kw)/(KₐKb)) Degree of hydrolysis is independent of concentration.

Answer 1

The Correct Option is A

Step 1: For a salt of weak acid and weak base: Kh=(Kw)/(KₐKb)
Step 2: Kh=frac10⁻14(1.8×10⁻5)(1.8×10⁻5) =3.09×10⁻5
Step 3: Degree of hydrolysis: h=√(Kh)=√3.09×10⁻5 =5.56×10⁻3
Step 4: Percentage hydrolysis: %h = h×100 = 0.556%