Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0184 M solution. The pH of the resulting solution is 2.32. Calculate the Ka for the acid.

Answer 1

use:
pH = -log [H+]
2.32 = -log [H+]
[H+] = 4.786*10^-3 M
HA dissociates as:

HA          ----->     H+   + A-
1.84*10^-2                 0         0
1.84*10^-2-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 4.786*10^-3*4.786*10^-3/(0.0184-4.786*10^-3)
Ka = 1.683*10^-3
Answer: 1.68*10^-3