tolong bantu jawab terimakasih

400 mL CH3COOH 0,5 M + 100 mL CH3CONa 0,5 M
Ka = 10^-5

pH ... ?

n CH3COOH = M × V
n CH3COOH = 0,5 × 400
n CH3COOH = 200 mmol

n CH3COONa = M × V
n CH3COONa = 0,5 × 100
n CH3COONa = 50 mmol

[H^+] = Ka × n asam lemah/n basa konjugasi
[H^+] = 10^-5 × 200/50
[H^+] = 10^-5 × 4

pH = - log [H^+]
pH = - log 10^-5 × 4
pH = 5 - log 4