Jika larutan asam asetat mempunyai pH=3 dan Ka=10 pangkat-5 (Mr=60),maka jumlah asam asetat dalam 1 liter larutan asam asetat sebesar...gram a.0.6 gr b.0.3 gr c.6gr d.3gr e.60gr (sama cara yaa...)
Dzaky111
PH = 3 ⇒ [H+] = 0,001 M [H+]² = Ka.[CH3COOH] [CH3COOH] = [H+]²/Ka = (10^-3)²/10^-5 = 10^-1 M masa CH3COOH = M x Mr = 0,1 x 60 = 6 gram
[H+]² = Ka.[CH3COOH]
[CH3COOH] = [H+]²/Ka
= (10^-3)²/10^-5 = 10^-1 M
masa CH3COOH = M x Mr
= 0,1 x 60
= 6 gram