Jawabanmu

2014-01-15T12:40:15+07:00
P[OH-] = -log [OH-]
14-12 = -log [OH-]
2 = -log [OH-]
2-log1 = - log[OH-]
1.10^-2=[OH-]
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2014-01-15T12:53:15+07:00
POH + pH = pKw
pOH + 12 = 14
pOH = 2
pOH = -log [OH-]
2=- log [OH-]
[OH-]=10^-2 M


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