Jawabanmu

2014-04-20T11:56:25+07:00
NaOH + CH3COOH ---->> CH3COONa + H2O
[H+] = √ka . Ma
 = 1,8 . 10^5 . 0,1
= 1,8. 10^6
PH = - log [H+]
PH = 6 - log 1,8
2014-04-20T14:55:48+07:00
NaOH + CH3COOH ---->> CH3COONa + H2O
[H+] = √ka . Ma
 = 1,8 . 10^5 . 0,1
= 1,8. 10^6
PH = - log [H+]
PH = 6 - log 1,8