Consider a titration of 24.2 mL of 0.0878M solution of aqueous ammonia [Kb(NH3) = 1.8e-5] with a 0.115M of hydrochloric acid.
Calculate:

a) the pH of the aqueous ammonia solution before the titration

b) the pH of the solution at half-equivalence point

c) the pH of the solution at the equivalence point

For part "a" I have:
[OH-] = √(1.8e-5 x 0.115)

For part "b" I have:

pOH = pKb

pOH = -log(1.8e-5)

then I subtract pOH from 14 to get pH