# The concentration of carbonic acid in normal blood is 0.029 mol/L and the pKa for carbonic acid is 6.1. Using the Henderson-Hasselbach equation, how do you determine the concentration of bicarbonate ion in normal blood sample which is pH 7.40?

Jun 11, 2016

$\left[{\text{HCO}}_{3}^{-}\right]$

#### Explanation:

All you have to do here is plug in the values given to you in the Henderson - Hasselbalch equation and solve for the concentration of the bicarbonate anions, ${\text{HCO}}_{3}^{-}$.

Now, let's assume that you're not familiar with the H - H equation. You can derive it by using the equilibrium reaction that describes the partial ionization of carbonic acid, ${\text{H"_2"CO}}_{3}$, in aqueous solution

${\text{H"_ 2 "CO"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "HCO}}_{3 \left(a q\right)}^{-}$

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \left(\left[{\text{H"_3"O"^(+)] * ["HCO"_3^(-)])/(["H"_2"CO}}_{3}\right]\right)$

Take the log base $10$ of both sides to get

log(K_a) = log( ["H"_3"O"^(+)] * (["HCO"_3^(-)])/(["H"_2"CO"_3]))

This will be equivalent to

log(K_a) = log(["H"_3"O"^(+)]) + log((["HCO"_3^(-)])/(["H"_2"CO"_3]))

Rearrange to get

- log(["H"_3"O"^(+)]) = - log(K_a) + log((["HCO"_3^(-)])/(["H"_2"CO"_3]))

But you know that

color(purple)(|bar(ul(color(white)(a/a)color(black)("pH" = - log(["H"_3"O"^(+)]))color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("p"K_a = - log(K_a))color(white)(a/a)|)))

and so you have

color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log((["HCO"_3^(-)])/(["H"_2"CO"_3]))color(white)(a/a)|))) -> the Henderson - Hasselbalch equation

This is the Henderson - Hasselbalch equation for buffer solutions that contain a weak acid and its conjugate base.

To find the concentration of bicarbonate anions, isolate the log term on one side of the equation

$7.40 = 6.1 + \log \left(\left(\left[{\text{HCO"_3^(-)])/(["H"_2"CO}}_{3}\right]\right)\right)$

$\log \left(\left(\left[{\text{HCO"_3^(-)])/(["H"_2"CO}}_{3}\right]\right)\right) = 1.3$

This will be equivalent to

${10}^{\log} \left(\left(\left[{\text{HCO"_3^(-)])/(["H"_2"CO}}_{3}\right]\right)\right) = {10}^{1.3}$

which in turn will get you

$\left(\left[{\text{HCO"_3^(-)])/(["H"_2"CO}}_{3}\right]\right) = 19.95$

Therefore, you will have

$\left[{\text{HCO"_3^(-)] = 19.95 * ["H"_2"CO}}_{3}\right]$

["HCO"_3^(-)] = 19.95 * "0.029 mol L"^(-1) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.58 mol L"^(-1))color(white)(a/a)|)))

Notice that you have a higher concentration of conjugate base than you do of weak acid, which is why the pH of the buffer is higher than the $\text{p} {K}_{a}$ of the weak acid.