what percentage of the acid is not ionized?

Question

A weak acid, HA, has a pK_a  of  4.810. If a solution of this acid has a pH of  4.763, what percentage of the acid is not ionized? Assume all  in the solution came from the ionization of HA.

Answer 1

Two ways to approach this:

pH = pKa + log [A-]/[HA]

4.763 = 4.810 + log [A-]/[HA]

log [A-]/[HA] = -0.047

[A-]/[HA] = 0.897 x100 = 89.7% ionized

Percent not ionized = 10.3%

Another way would be:

HA <==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA] and since pKa = 4.810, the Ka = 1x10^-4.810 = 1.55x10^-5

We know the pH = 4.763, so we can find the [H⁺] and [A^-]

pH = -log [H⁺]

[H⁺] = 1x10^-4.763 = 1.73x10^-5 M

We can now find the initial [HA]:

1.55x10^-5 = (1.73x10^-5)(1.73x10^-5)/[HA] - 1.73x10^-5 (I will omit the [H⁺] in the denominator to make it easier)

2.99x10^-10 = 1.55x10^-5[HA]

[HA] = 1.93x10^-5

% ionization = 1.73/1.99 (x100%) = 86.9%

% non ionized = 13.1%

This is close to the first method, but varies based on neglecting the 1.73x10^-5 in the denominator.