Jika tetapan asam CH3COOH=10-5, maka pH larutan CH3COONa 0,1 M adalah...

[OH-] = √Kw/Ka . [anion]
[OH-] = √10^-14 / 10^-5 . 10^-1
[OH-] = √10^-9 . ^10^-1
[OH-] = √10^-10
[OH-] = 10^-5

pOH = -log [OH-]
pOH = -log 10^-5
pOH = 5

pH = 14 - 5
pH = 9

→ Atau Menggunakan Cara :
pH = 1/2 (14 + pKa + log[G])
pH = 1/2 (14 + (-log10^-5) + log10^-1)
pH = 1/2 (14 + 5 -1)
pH = 1/2 (18)
pH = 9