7.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

${\mathrm{NaNO}}_{2}$is the salt of a strong base (NaOH) and a weak acid (${\mathrm{HNO}}_{2}$).

${\mathrm{NO}}_{2}^{-}+{\mathrm{H}}_{2}\mathrm{O}↔{\mathrm{HNO}}_{2}+{\mathrm{OH}}^{-}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{HNO}}_{2}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NO}}_{2}^{-}\right]}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{\alpha }}}=\frac{{10}^{-14}}{4.5×{10}^{-4}}=.22×{10}^{-10}$

Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species present in the solution will be: