tolong di jawab ya kakak terimakasih

0,1 mol CH3COOH + 0,1 mol Mg(CH3COO)2
V = 1 Liter.
Ka = 2 × 10^-5

pH ... ?

[H^+] = Ka × mol Asam lemah/mol B. konjugasi
[H^+] = 2 × 10^-5 × 0,1/0,1
[H^+] = 2 × 10^-5

pH = - log [H^+]
pH = - log 2 × 10^-5
pH = 5 - log 2