Question

Carbonate buffers are important in regulating the pH of the blood at 7.40. If the carbonic acid concentration in a sample of blood is 0.0012 M, determine the bicarbonate ion concentration required to buffer the pH of blood atpH = 7.40.
${\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{H}\mathit{C}{{\mathit{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}{\mathit{K}}_{\mathbf{a}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$

Short answer

The bicarbonate ion concentration required to buffer the pH of the blood at pH = 7.4 is$1.3\times {10}^{-2}.$

Step-by-step solution

pH of blood buffer and concentration of carbonic acid

The pH of the blood buffer is 7.4 whereas the concentration of carbonic acid in the blood sample is 0.0012 M.

$K_{a}of{H}_{2}C{O}_3=4.3\times {10}^{-7}$

Using the Henderson-Hasselbalch equation to calculate the concentration of HCO3

The concentration can be calculated as,

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}\\ 7.4=-\log (4.3\times {10}^{-7})+\log \frac{\left[HC{O}_3^{-}\right]}{\left[0.0012\right]}\\ \log \frac{\left[HC{O}_3^{-}\right]}{\left[0.0012\right]}=7.4-6.37\\ \log \frac{\left[HC{O}_3^{-}\right]}{\left[0.0012\right]}=1.03\\ \log \frac{\left[HC{O}_3^{-}\right]}{\left[0.0012\right]}={10}^{1.03}\\ \left[HC{O}_3^{-}\right]=1.3\times {10}^{-2}\end{array}$