The expression to be used to calculate the pH of sodium bicarbonate solution is

where $K_{\mathrm{a}1}^{\circ }$ and $K_{\mathrm{a}2}^{\circ }$  stand for the standard ionization constants of ${\mathrm{H}}_2{\mathrm{CO}}_3^{-}$ and  ${\mathrm{HCO}}_3^{-}$, respectively.

$\begin{array}{l}{\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{OH}}^{-};K_{\mathrm{h}}=\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a}1}}\\ {\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CO}}_3^{2-};K_{\mathrm{a}2}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}\\ \frac{K_{\mathrm{h}}}{K_{\mathrm{a}2}}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a}1}{K}_{a2}}=\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{CO}}_3^{-}\right]}\frac{\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\simeq \frac{K_{\mathrm{w}}}{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2}\end{array}$

(Since $\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right];\left[{\mathrm{CO}}_3^{2-}\right]$)

Also, $K_{\mathrm{h}}/K_{\mathrm{a}2}=K_{\mathrm{w}}/K_{\mathrm{a}1}{K}_{\mathrm{a}2}$ Hence ${\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2=K_{\mathrm{a}1}{K}_{\mathrm{a}2}$ or $\mathrm{pH}=\frac{1}{2}\left(\mathrm{p}{K}_{\mathrm{a}1}^{\circ }+\mathrm{p}{K}_{\mathrm{a}2}^{\circ }\right)$