The pH of 0.1 M NaHCO₃ solution can be computed from the expression

Na⁺ does not show hydrolysis reaction. HCO₃ is an amphiprotic anion

${\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CO}}_3^{2-};K_{\mathrm{a}2}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}$

${\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{OH}}^{-};K_{\mathrm{b}}=\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}$

also $K_{\mathrm{b}}=\frac{\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]\left[{\mathrm{H}}^{+}\right]/\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=\frac{K_{\mathrm{w}}}{K_{\mathrm{al}}}$

From the expression of K_a2 ,we get

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{K_{\mathrm{a}2}\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{CO}}_3^{2-}\right]}=\frac{K_{\mathrm{a}2}\left\{\left[{\mathrm{H}}_2{\mathrm{CO}}_3^{-}\right]\left[{\mathrm{OH}}^{-}\right]/K_{\mathrm{b}}\right\}}{\left[{\mathrm{CO}}_3^{2-}\right]}=\frac{K_{\mathrm{a}2}}{\left[{\mathrm{CO}}_3^{2-}\right]}\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\left(K_{\mathrm{w}}/K_{\mathrm{a}1}\right)}\frac{K_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$or  ${\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2=K_{\mathrm{a}1}{K}_{\mathrm{a}2}\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{CO}}_3^{2-}\right]}$

Since both K_a2 and K_b are usually very small, one may assume $\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]=\left[{\mathrm{CO}}_3^{2-}\right]$ Hence${\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2=K_{\mathrm{a}1}{K}_{\mathrm{a}2}$ or $\mathrm{pH}=\frac{1}{2}\left(\mathrm{p}{K}_{\mathrm{a}1}^{\circ }+\mathrm{p}{K}_{\mathrm{a}2}^{\circ }\right)$