Explanation

(a)
$aX + H_{2}O \rightleftharpoons NaOH + HX$
In it $HX$ is weak acid, so $NaX$ is a salt of weak acid and strong base.
$\therefore$ Hydrolysis constant of $NaX$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{1 \times 10^{- 14}}{10^{- 5}} = 1 \times 10^{- 9}$
For this type of salt
$K_{h} = \frac{h^{2}}{V} = Ch^{2}$
$h = degree\ of\ hydrolysis$
$\therefore$ $1 \times 10^{- 9} = 0.1 \times h^{2}$
or $h^{2} = \frac{1 \times 10^{- 9}}{0.1} = 1 \times 10^{- 8}$
$h = 1 \times 10^{- 4}$
$\therefore\ \%$ of degree of hydrolysis on $NaX$ salt
$= 1 \times 10^{- 4} \times 100 = 1 \times 10^{- 2} = 0.01\%$