Sodium nitrite is a salt of NaOH (strong base) and HNO_{2} (weak acid).

\( NO_{2}^{-} + H_{2}O \leftrightarrow HNO_{2} + OH^{-} \)

\( K_{h} = \dfrac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]} \)

\( \Rightarrow \;\dfrac{K_{w}}{K_{a}} \)

\( = \dfrac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}\)

Let, y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:

\( [NO_{2}^{-}] = 0.04 – y; 0.04 \)

\( [HNO_{2}] = y\)

\( [OH^{-}] = y\)

\( K_{h} = \dfrac{y^{2}}{0.04} = 0.22\times 10^{-10} \)

\( y^{2} = 0.0088\times 10^{-10} \)

\( y = 0.093\times 10^{-5}y \)

\( [OH^{-}]= 0.093\times 10^{-5}M\)

\( [H_{3}O^{+}] = \dfrac{10^{-14}}{0.093\times 10^{-5}} \)

\( = 10.75\times 10^{-9}\)

Thus, \( pH = -\log (10.75\times 10^{-9})= 7.96\)

Thus, the degree of hydrolysis is

\( = \dfrac{y}{0.04} = \dfrac{0.093\times 10^{-5}}{0.04} \)

= \( 2.325\times 10^{-5}\)

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