Jawabanmu

2014-03-11T10:59:03+07:00
Ca(OH)2 termasuk dalam basa kuat
sehingga [OH-] = 2 x 0,03 = 0,06 M = 6 x 10^-2
pOH = 2 - Log 6
pH = 14-(2-Log6)
= 12+log6
2014-03-11T11:15:44+07:00
Dik : vol. Ca(OH)2 = 200 mL = 0,2 liter
Molaritas Ca(OH)2 = 0,03 M
Dit : pH = ....?
Jawab :
reaksi ionisasi : Ca(OH)2(aq)-> Ca2+(aq) + 2 OH–(aq)
[OH^-] = koefisien atau jumlah ion OH^- x Molaritas
[OH^-] = 2 x 0,03
[OH^-] =  6 x 10^-2

pOH = -log [OH^-]
pOH = - log 6 x 10^-2
pOH = 2 - log 6
pOH = 2 - 0,448
pOH = 1,222

pH = 14 - pOH
pH = 14 - 1,222
pH = 12,778