suatu campuran sebanyak 100 ml terdiri dari larutan CH3COOH 0,01 mol dan CH3COONa 0,02 mol ( Ka CH3COOH=10-5

H+=ka x mol asam/garam=10-5.0,01/0,02=5.10-6
pH=6-log5

Diketahui :
- (a) jumlah mol CH3COOH = 0,01 mol
- (g) jumlah mol CH3COONa = 0,02 mol
- ka CH3COOH = 10^-5
Ditanya : PH?
Jawab :
[H+] = ka x a/g
= 10^-5 x 0,01/0,02
= 5 x 10^-6

PH = - log [H+]
= - log 5 x 10^-6
= 6 - log 5