Question

If a buffer solution is 0.140 M in a weak base (Kb = 5.3 × 10-5) and 0.530 M in its conjugate acid, what is the pH?

Answer 1

B + H2O <--> BH+ + OH-
Kb = [BH+][OH-]/[B] = 1.0 X 10^-5
Substituting the concentrations of the base and its conjugate acid gives:
1.0 X 10^-5 = (0.410) [OH-] / (0.140)
[OH-] = 3.41 X 10^-6
From this, and the expression for Kw,
[H3O+] = 1.0 X 10^-14 / 3.41 X 10^-6 = 2.9X10^-9
and pH = - log 2.9X10^-9 = 8.53
You can also use Kb to calculate Ka and the pKa, and then use the Henderson-Haselbalch equation:
Ka X Kb = Kw
Ka = 1X10^-14 / 1X10^-5 = 1.0 X 10^-9
pKa = 9.00
Then, pH = pKa + log [base]/[acid]
pH = 9.00 + log 0.14 / 0.41 = 8.53