Question

If the K_a of a monoprotic weak acid is 5.6 times 10^-6, what is the pH of a 0.46 M solution of this acid? pH = 6.79

Answer 1

Let the monoprotic acid be HA.

HA dissociates as:

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

0.46 0 0

0.46-x x x

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

Ka = x*x/(c-x)

since ka is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.6*10^-6)*0.46) = 1.605*10^-3

use:

pH = -log [H+]

= -log (1.605*10^-3)

= 2.79

Answer: 2.79