Question

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?

Answer 1

`["HCO"_3^(-)]`

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, `"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, `"H"_2"CO"_3`, in aqueous solution

`"H"_ 2 "CO"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "HCO"_ (3(aq))^(-)`

By definition, the acid dissociation constant, `K_a`, will be equal to

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

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((["HCO"_3^(-)])/(["H"_2"CO"_3]))`

`log((["HCO"_3^(-)])/(["H"_2"CO"_3])) = 1.3`

This will be equivalent to

`10^log((["HCO"_3^(-)])/(["H"_2"CO"_3])) = 10^1.3`

which in turn will get you

`(["HCO"_3^(-)])/(["H"_2"CO"_3]) = 19.95`

Therefore, you will have

`["HCO"_3^(-)] = 19.95 * ["H"_2"CO"_3]`

`["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 `"p"K_a` of the weak acid.