Question

A 0.282-M aqueous solution of a weak acid has a pH of 2.81. Calculate Ka for the acid.

Answer 1

use:
pH = -log [H+]
2.81 = -log [H+]
[H+] = 1.549*10^-3 M

Let the weak acid be written as HA
HA dissociates as:

HA          ----->     H+   + A-
0.282                 0         0
0.282-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.549*10^-3*1.549*10^-3/(0.282-1.549*10^-3)
Ka = 8.55*10^-6
Answer: 8.55*10^-6