Question

The percentage degree of hydrolysis in the equilibrium at salt concentration $0.0001M$ is :
$A^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HA\left(aq\right)+O{H}^{-}\left(aq\right)$

$(\text{Given}{K}_a=1.0\times {10}^{-6})$

Answer 1

Given,
$C={10}^{-4}M{K}_a={10}^{-6}$
The relationship between hydrolysis constant and dissociation constant is given as:
$K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}$
The expression for degree of hydrolysis is:
$h=\sqrt{\frac{K_h}{C}}$
$\Rightarrow h=\sqrt{\frac{{10}^{-8}}{{10}^{-4}}}=\sqrt{{10}^{-4}}={10}^{-2}$
Percentage of degree of hydrolysis $={10}^{-2}\times 100\%=1\%$