Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0122 M solution. The pH of the resulting solution is 2.64. Calculate the Ka for the acid.

Answer 1

concentration = 0.0122 M

pH of solution = 2.64

pH = -log [H+]

[H+] = 10^-2.64

[H+] = 2.29 x 10^-3 M

           HA     -------------------->   H+ +      A-

0.0122 - 2.29 x 10^-3                     2.29 x 10^-3 2.29 x 10^-3

Ka = [H+][A-] / [HA]

     = (2.29 x 10^-3)^2 / 0.0122 - 2.29 x 10^-3

     = 5.29 x 10^-4

Ka of acid = 5.29 x 10^-4