Question

If the Ka of a monoprotic weak acid is 7.5 × 10-6, what is the pH of a 0.12 M solution of this acid?

Answer 1

Lets write the acid as HA

Lets write the dissociation equation of HA

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

0.12 0 0

0.12-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((7.5*10^-6)*0.12) = 9.487*10^-4

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

so, x = 9.487*10^-4 M

so,

[H+] = x = 9.487*10^-4 M

we have below equation to be used:

pH = -log [H+]

= -log (9.487*10^-4)

= 3.02

Answer: 3.02