1. Berapa pH larutan yang terjadi jika dalam 500 mL larutan terdapat 12 gram CH₃COOH dan 8,2 gram CH₃COONa? (Ka=1,8x10⁻⁵)

12 gram CH3COOH + 8,2 gram CH3COONa
V = 500 mL
Ka = 1,8 × 10^-5
Mr CH3COOH = 60
Mr CH3COONa = 82

pH larutan ... ?

n CH3COOH = gram/Mr
n CH3COOH = 12/60
n CH3COOH = 0,2 mol

n CH3COONa = gram/Mr
n CH3COONa = 8,2/82
n CH3COONa = 0,1 mol

[H^+] = Ka × n Asam lemah/n B. konjugasi
[H^+] = 1,8 × 10^-5 × 0,2/0,1
[H^+] = 1,8 × 10^-5 × 2
[H^+] = 3,6 × 10^-5

pH = - log [H^+]
pH = - log 3,6 × 10^-5
pH = 5 - log 3,6