Jawabanmu

2014-03-06T11:25:11+07:00
[OH]=akar Kb x M
       =akar 1x10^-6 x 0,01
       =akar 1x10^-6 x 10^-2
       =akar 1x10^-8
       =1x10^-4
pOH=-log[OH] = -log 1x10^-4 = 4
pH= 14-pOH = 14-4=10