Jawabanmu

2014-04-30T05:39:42+07:00
[H+]=√(kw/kb)x[garam] =√(10^-14/1,8.10^-5)x[50ml.0,1M/50+50] =√5,5.10^-10 x 5.10^-2 =√2,7.10^-11 =√5,2 x 10^-6 pH=-log [H+ ] =-log[ √5,2 x 10^-6] = 6-log √5,2