SAMPLE PROBLEM 18.10
(continued)
[Ac-] = 0.25 M - x 0.25 M (since Kb is small)
5.6 x 10-10 x2/0.25 M
x 1.2 x 10-5 M = [OH-]
pH = -log (8.3 x 10-10 M) = 9.08