sebanyak 1 liter larutan CH3COOK mempunyai pH = 8 . jika Ka CH3COOH = 10^-5 dan Mr CH3COOK =98, massa garam CH3COOK yang terlarut dalam 1 liter larutan CH3COOK

pOH = 14 - 8 = 6
pOH = - log OH-
6 = -log OH-
OH- 10^-6

OH- = √ kw / ka x Mg
10^-6 = √ 10^-14 / 10^-5 x Mg
10^-12 = 10^-9 x Mg
Mg = 10^-3 M = 0.001 M

Mg = n / V
0.001 M = n / 1 L
n = 0.001 mol

massa = n x Mr
massa = 0.001 x 98
massa = 0.098 gr = 98 mg