Question

Calculate the % hydrolysis of 0.1 M $C{H}_{3}COON{H}_4$, when $K_a\left(C{H}_{3}COOH\right)=K_b\left(N{H}_{4}OH\right)=1.8\times {10}^{-5}$.

• A. $0.55$
• B. $7.63$
• C. $0.55\times {10}^{-2}$
• D. $7.63\times {10}^{-3}$

Answer 1

The correct option is A $0.55$

$C{H}_{3}COO{H}_4$ is a salt of weak acid and weak base.

$K_h=\frac{K_w}{K_a\times K_b}$

we know that $K_h=\frac{{\alpha }^2}{(1-\alpha {)}^2},\alpha =\%$ of hydrolysis.

Dissociation is independent upon concentration.

$K_b=K_a=1.8\times {10}^{-5}$

Thus,

$\frac{{\alpha }^2}{(1-\alpha {)}^2}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}$

or $\frac{\alpha }{1-\alpha }=\sqrt{\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}}$

$\frac{\alpha }{1-\alpha }=\frac{{10}^{-2}}{1.8}=\frac{1}{180}$

$\therefore \alpha =0.55\times {10}^{-2}$

Hence, option C is correct.