jika Ka CH3COOH = 10 pangkat -5 maka pH larutan CH3COONa 0,9 M adalah

[OH-] = √Kw / Ka . n . M
[OH-] = √10^-14 / 10^-5 . 9 x 10^-1
[OH-] = √10^-9 . 9 . 10^-1
[OH-] = √9 x 10^-10
[OH-] = 3 x 10^-5

pOH = -log[OH-],
pOH = -log 3 x 10^-5
pOH = 5 - log3

pH = 14 - (5 - log3)
pH = 14 - 5 + log3
pH = 9 + log3

→ Atau :

pH = 1/2 (14 + pKa + log[G])
pH = 1/2 (14 + (-log10^-5) + log9x 10^-1)
pH = 1/2 (14 + 5 - 1 + log9)
pH = 1/2 (18 + log9)
pH = 9 + 1/2log9