# How would you use the Henderson-Hasselbalch equation to calculate the ratio of H2CO3 to HCO3- in blood having a pH of 7.40?

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{{A}^{-}}{H A}\right\}$. Clearly, we need $p {K}_{a}$ for ${H}_{2} C {O}_{3}$, i.e. $p {K}_{a} = 6.35 .$
If $p H = 7.40$, then ${\log}_{10} \left\{\frac{H C \left(= O\right) {O}^{-}}{H C \left(= O\right) O H}\right\}$ $=$ $1.05$. This reasonable because the $p H$ is higher than the $p {K}_{a}$ and there should be more carbonate than carbonic acid.
Thus $\left\{\frac{H C \left(= O\right) {O}^{-}}{H C \left(= O\right) O H}\right\}$ $=$ ${10}^{1.05}$ $=$ ??