What percentage of the carbon in the ${\mathrm{H}}_2{\mathrm{CO}}_3-{\mathrm{HCO}}_3^{-}$ buffer should be in the form of ${\mathrm{HCO}}_3^{-}$ so as to have a neutral solution at 25° C? $\left(K_a=4\times {10}^{-7}\right)$

As a buffer, using the Henderson-Hasselbalch equation, $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{{\mathrm{A}}^{-}}{\mathrm{HA}}$

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}--\left(1\right)\\ {\mathrm{pK}}_{\mathrm{a}}=-\log \left({\mathrm{K}}_{\mathrm{a}}\right)=-\log \left(4\times {10}^{-7}\right)=-\log 4+7\\ \mathrm{pH}=7\text{ (Neutral solution) }\end{array}$

Replace 1 with:

$\begin{array}{l}7=-\log 4+7+\log \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}\\ \log \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=\log 4\\ \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=\frac{4}{1}\end{array}$

Therefore, the shape of the buffer's percentage ${\mathrm{HCO}}_3^{-}=\frac{4}{5}\times 100=80\mathrm{\%}$.