# How is the Henderson-Hasselbalch equation used to calculate the ratio of H_2CO_3 to HCO_3^- in blood having a pH of 7.40?

May 29, 2017

$\left(\left[H C {O}_{3}^{-}\right] : \left[{H}_{2} C {O}_{3}\right]\right) = 10 : 1$

#### Explanation:

$p H = p K a + {\log}_{10} \left(\frac{\left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}\right)$

$K {a}_{1} \left({H}_{2} C {O}_{3}\right) = 4.5 x {10}^{- 7}$
=> $p K a = - \log {K}_{a}$ = $- \log \left(4.5 x {10}^{- 7}\right)$ =$6.4$

Given $p H = 7.4$

Substitute into HH Equation and solve for $\left(\frac{\left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}\right)$

$7.4 = 6.4 + {\log}_{10} \left(\frac{\left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}\right)$

${\log}_{10} \left(\frac{\left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}\right) = 7.4 - 6.4 = 1.0$

$\left(\frac{\left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}\right) = {10}^{1.0} = 10$

=> $\left(\left[H C {O}_{3}^{-}\right] : \left[{H}_{2} C {O}_{3}\right]\right) = 10 : 1$