Question

The Ka of a weak acid is 1.73 x 10-5. Assuming the "x is small" approximation is valid, what is the predicted pH of a 0.10 M solution of the weak acid?

Answer 1

Let the weak acid be written as HA

HA dissociates as:

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

0.1 0 0

0.1-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.73*10^-5)*0.1) = 1.315*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.315*10^-3 M

so.[H+] = x = 1.315*10^-3 M

use:

pH = -log [H+]

= -log (1.315*10^-3)

= 2.88

Answer: 2.88