Explanation

(a)
Given, $K_{w} = 10^{- 14},\ K_{a} = 10^{- 5}$
Concentration of salt = 0.001 M
$\therefore$ $K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9}$
According to equation
$A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$
Let degree of hydrolysis=$h$
$\therefore\ \ \ \ \ \ 0.001(1 - h)(0.001 \times h)(0.001 \times h)$
$\therefore$ $K_{h} = \frac{\lbrack HA\rbrack\lbrack OH^{-}\rbrack}{\lbrack A^{-}\rbrack} = \frac{(0.001 \times h)(0.001 \times h)}{0.001(1 - h)}$
or $10^{- 9} = (0.001h)^{2}$ $\lbrack\because 0.001(1 - h) = 1\rbrack$
or $10^{- 6} = h^{2}$
$\therefore$ $10^{- 3} = h$