Jawabanmu

2014-06-17T16:14:29+07:00
Mol CH3COOH =50ml x 0,1 M = 5 mmol
mol CH3COONa = 50ml x 0,1 M = 5mmol
[H⁺]= Ka.mol as/mol garam
       = 1,8x10⁻⁵ x 5/5
       =1,8x 10⁻⁵
pH = 5-log1,8
2014-06-17T16:18:15+07:00
H⁺ = ka. (na/ng)
= 1,8 . 10⁻⁵ . (5/5)
= 1,8 . 10⁻⁵
pH = 5 - log 1,8