How do I solve this AP Chemistry problem on Kb

Question

In the laboratory, a general chemistry student measured the pH of a 0.599M aqueous solution of C2H5NH2 (a weak base) to be 12.192. What is Kb for this base?

Answer 1

C2H5NH2 + H2O ==>C2H5NH3⁺ + OH^-

Kb = [C2H5NH3⁺][OH^-]/[C2H5NH2] 

From pH = 12.192, one can calculate pOH, and then OH^-.

pOH = 14-pH = 14 -12.192 = 1.808

[OH-] = 10^-1.808 = 1.556x10^-2 M

Kb = (1.556x10^-2)(1.556x10^-2)/0.599

Kb = 4.042x10^-4