Verified answer

This problem is worked similarly to the NH3/HNO3 question you asked:

Equilibrium: CH3COOH <--> H+ + CH3COO-

Ka = 1.8X10^-5 = [H+][CH3COO-]/[CH3COOH]

Initial Moles CH3COOH = 0.052 L X 0.350 mol/L = 0.0182 mol

Moles NaOH added = 0.0230 L X 0.400 mol/L = 9.2X10^-3

NaOH neutralizes CH3COOH quantitatively forming 9.2X10^-3 mol CH3COO-

Moles CH3COOH remaining = 0.0182 - 0.0092 = 9.00X10^-3 mol CH3COOH

Ka = 1.8X10^-5 = [H+] (9.0X10^-3) / (9.2X10^-3)

[H+] = 1.84X10^-5 M

pH = 4.74