Question

If the Ka of a monoprotic weak acid is 8.2 × 10-6, what is the pH of a 0.32 M solution of this acid?

Answer 1

monoprotic acid (HA) = 0.32 M

HA + H2O   ----------------> A-   + H3O+

0.32                                     0            0

0.32 - x                                  x          x

Ka = x^2 / 0.32 - x

8.2 x 10^-6 = x^2 / 0.32 - x

x = 1.62 x 10^-3

[H3O+] = 1.62 x 10^-3 M

pH = -log [H3O+] = -log (1.62 x 10^-3 )

pH = 2.8