Question

If a buffer solution is 0.140 M in a weak acid (Ka = 1.5 × 10-5) and 0.560 M in its conjugate base, what is the pH?

Answer 1

Since you are given K_b, the equilibrium between the weak base, water and its conjugated acid is:

B + H₂O <--> BH⁺ + OH^-

K_b = [BH⁺][OH^-]/[B] = 1.0 X 10^-5

Substituting the concentrations of the base and its conjugate acid gives:
1.0 X 10^-5 = (0.560) [OH^-] / (0.140)
[OH^-] = 2.5 X 10^-6

From this, and the expression for K_w,
[H₃O⁺] = 1.0 X 10^-14 / 2.5 X 10^-6 = 4X10^-9

and pH = -log 4X10^-9 = 8.40