Question

The degree of hydrolysis of $0.15M$ solution of ammonium acetate is:

[$K_a$ for ${CH}_{3}COOH$ is $1.8\times {10}^{-5}$ and $K_b$ for ${NH}_3$ is $1.8\times {10}^{-9}$]

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

Answer 1

The correct option is A $0.556$

$K_a$ for ${CH}_{3}COOH$ is $1.8\times {10}^{-5}$ and $K_b$ for ${NH}_3$ is $1.8\times {10}^{-5}$.
The degree of hydrolysis
$h=\sqrt{\frac{K_w}{K_a\times K_b}}=\sqrt{\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-9}}}=0.556$.
The degree of hydrolysis of $0.15M$ solution of ammonium acetate is $0.556$.
Note: The degree of hydrolysis of salt of the weak acid and weak base is independent of its concentration.