The ionization constant of nitrous acid is $4.5 \times 10^{-4}$. Calculate the $pH$ of $0.04 \ M$ sodium nitrite solution and also its degree of hydrolysis.

$NaNO_{2}$ is the salt of a strong base $(NaOH)$ and a weak acid $(HNO_{2})$.
$NO_{2}^{-} + H_{2}O \longleftrightarrow HNO_{2} + OH^{-}$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}$
For salt hydrolysis,$K_{h} = \frac{h^{2}C}{1-h} \approx h^{2}C$ (since $h$ is very small).
$h = \sqrt{\frac{K_{h}}{C}} = \sqrt{\frac{2.22 \times 10^{-11}}{0.04}} = \sqrt{5.55 \times 10^{-10}} = 2.356 \times 10^{-5}$.
$[OH^{-}] = C \times h = 0.04 \times 2.356 \times 10^{-5} = 9.424 \times 10^{-7} \ M$.
$pOH = -\log(9.424 \times 10^{-7}) = 7 - \log(9.424) = 7 - 0.974 = 6.026$.
$pH = 14 - pOH = 14 - 6.026 = 7.974$.