Question:medium
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:
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:
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For salts of weak acid and weak base:
h=√((Kw)/(KₐKb))
Degree of hydrolysis is independent of concentration.
Updated On: Mar 19, 2026
- \(0.556\)
- \(4.72\)
- \(9.38\)
- 5.56
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The Correct Option is A
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