Question

.$\%$ hydrolysis of $0.1M{CH}_{3}COO{NH}_4$ when $K_a=K_b=1.8\times {10}^{-5}$ is?

Answer 1

For hydrolysis of salt of weak acid and weak base,

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

$K_w={10}^{-14}$

$K_a=K_b=1.8\times 0^{-5}\left(\text{Given}\right)$

$\therefore K_H=\frac{{10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}$

$\Rightarrow K_H=0.3\times {10}^{-4}$

As we know that,

$K_H=\frac{h^2}{1-h^2}$

$0.3\times {10}^{-4}=\frac{h^2}{1-h^2}$

$\Rightarrow (1.3\times {10}^{-4})h^2=0.3\times {10}^{-4}$

$\Rightarrow h^2=\frac{0.3}{1.3}$

$\Rightarrow h=\sqrt{0.23}=0.48$

$\%\text{hydrolysis}=0.48\times 100=48\%$

Hence the required answer is $48\%$.