A 0.411 M aqueous solution of weak monoprotic acid (HA) has a pH of 4.504, what is the pKa of the acid dissociation constant?

Question

Please explain how to calculate pKa.

Answer 1

pKa is the negative log of the Ka. So to answer this question, we first need to find the Ka.

A weak monoprotic acid can be written simply as HA.

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

From the pH = 4.504, we can find the [H+] which will also be the [A-]

[H⁺] = [A^-] = 1x10^-4.504 = 3.13x10^-5 M

Ka = (3.13x10^-5)(3.13x10^-5) / 0.411

Ka = 2.384x10^-9

pKa = -log 2.38x10^-9

pKa = 8.62