Determine the pH of a solution in which 300 mL of 01 M aqueo

Determine the pH of a solution in which 30.0 mL of 0.1 M aqueous acetic acid (K_a = 1.8 x 10^-5) has been titrated with 20.0 mL of 0.1 M NaOH.

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Answer: 8.95

CH3COOH + OH- <------> CH3COO- + H2O

pH = pka + log[Ch3COO-]/[CH3COOH] , pka = -logKa = -log(1.8 x10^ -5) = 4.75,

[CH3COOH] = ( 0.03x0.1-0.02x0.1)/(0.03+0.02) =0.02 , [CH3COO-] =(0.02x0.1)/0.03+0.02) = 0.04

pH = 4.75 + log(0.04/0.02)

= 5.05

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