Question

In the equilibrium, $A^{-}+H_{2}O\rightleftharpoons HA+O{H}^{-}$ $(K_a=1.0\times {10}^{-5})$, the degree of hydrolysis of 0.001 M solution of the salt is :

• A. ${10}^{-3}$
• B. ${10}^{-4}$
• C. ${10}^{-5}$
• D. ${10}^{-6}$

Answer 1

The correct option is A ${10}^{-3}$

The dissociation constant $K_a$ is $1.0\times {10}^{-5}$.

The relationship between the hydrolysis constant and dissociation constant is as given below.

$K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-5}}={10}^{-9}$.

The expression for the degree of hydrolysis is $h=\sqrt{\frac{K_h}{C}}$.

Substitute values in the above expression.

$h=\sqrt{\frac{{10}^{-9}}{0.001}}={10}^{-3}$.

Option A is correct.