Find pH of the solution?

Question

Given: Ka of acetic acid is 1.8x10-5, and initial volume and concentration of CH3COOH is 0.2M and 0.1L, total volume at the eqvivalence point is 0.2L.

Answer 1

0.1 litre @ 0.2 mol/litre = 0.02 moles =of acid , which produced 0.2 moles of acetate @ the equiv point

0.02 moles / 0.2 litre = 0.1 Molar acetate

acetate is the conjugate base of acetic acid :

Kb = Kwater / Ka = 1e-14 / 1.8e-5 = 5.56e-10

acetate in water -->  acetic acid & OH-

Kb = [acid] [OH-] / [acetate]

5.56e-10 = [x] [x] / [0.1 molar]

x = OH- = 7.45e-6

pOH = 5.13

pH = 14 - pOH

your answer is pH = 8.87