Question

A 0.50 M solution of a weak acid has a pH of 2.9. Calculate the Ka.

Answer 1

Let the weak acid be HA

use:
pH = -log [H3O+]
2.9 = -log [H3O+]
[H3O+] = 1.259*10^-3 M
HA dissociates as:

HA          ----->     H+   + A-
0.5                 0         0
0.5-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.259*10^-3*1.259*10^-3/(0.5-1.259*10^-3)
Ka = 3.18*10^-6
Answer: 3.18*10^-6