Question

Calculate the degree of dissociation of hydrolysis of a mixture containing 0.1N Nh4OH and 0.1N HCN. If Ka =10^-5 and Kb = 10^-5

Answer 1

Answer:

A value chain consisting of two types of additional storage in the matter of time

Answer 2

Answer:

The degree of hydrolysis of the mixture of $0.1N$ $NH_{4}OH$ and $0.1N$ $HCN$, $K_{h}$ = $10^{-2}$.

Explanation:

Given,

The degree of hydrolysis of $NH_{4}OH$, $K_{a}$ = $10^{-5}$

The degree of hydrolysis of $HCN$, $K_{b}$ = $10^{-5}$

The degree of hydrolysis of the mixture of $0.1N$ $NH_{4}OH$ and $0.1N$ $HCN$, $K_{h}$ =?

As we know,

• $NH_{4}OH$ is a weak base
• $HCN$ is a weak acid

Therefore, salt is a weak acid-weak base (WA-WB)

Now, for WA-WB, the degree of hydrolysis;

• $K_{h}$ = $\sqrt{\frac{K_{W} }{K_{a}\times K_{b} } }$

Here, $K_{W}$ = $10^{-14}$

• $K_{h}$ = $\sqrt{\frac{10^{-14} }{10^{-5} \times 10^{-5} } }$ = $\sqrt{10^{-4} }$ = $10^{-2}$

Hence, the degree of hydrolysis of the mixture of $0.1N$ $NH_{4}OH$ and $0.1N$ $HCN$, $K_{h}$ = $10^{-2}$