Hitunglah pH larutan garam jika 4,1 gram garam CH3COONa dilarutkan dalam air sampai volume 250ml (Ar C=12,H=1,O=16,Na=23). (Kh=2x10^-9)

Mr CH3COONa = 4 Ar C + 3 Ar H + 2 Ar O + Ar Na
Mr CH3COONa = 2(12) + 3(1) + 2(16) + 23
Mr CH3COONa = 82

n CH3COONa = massa / Mr
n CH3COONa = 4,1 gram / 82
n CH3COONa = 0,05 mol

V = 250 mL = 0,25 L

M = n / V
M = 0,05 mol / 0,25 L
M = 0,2 M

CH3COONa => CH3COO^{-} + Na^{+}
........0,2 M.................0,2 M............ 0,2 M

Garam ini merupakan Garam Basa karena terbuat dari asam lemah CH3COOH dan basa kuat NaOH
Maka cari [OH-]

[OH-] = √Kh × [CH3COO^{-}]
[OH-] = √ 2 × 10^-9 × 0,2 M
[OH-] = 2 × 10^-5

pOH = - log [OH-]
pOH = - log (2 × 10^-5)
pOH = 5 - log 2

pH = pKw - pOH
pH = 14 - (5 - log 2)
pH = 9 + log 2