The pH of blood is 7 .40. What is the ratio of $\frac{\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]}$ in the  blood? [pK_a (${\mathrm{H}}_2{\mathrm{PO}}_4^{-}$ ) = 7. 1 0]

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{PO}}_4^{-}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{HPO}}_4^{2-} \\ \mathrm{pH}=\mathrm{p}{K}_a+\log \frac{\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]} \\ 7.40=7.10+\log \frac{\left[\mathrm{HP}{O}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]} \\ 0.3=\log \frac{\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]} \\ \therefore \frac{\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]}=\frac{2}{1} \end{gathered}$$