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= akar 2X10 pangkat -5 X 0,2
=akar 4 X 10 pangkat -6
=2 x 10-3
pOH= 3-log 2
pH= 11+log 2
[OH-] = √Kb x M = √2.10-5 x 2. 10-1 = √4 x 10-6 = 2 x 10-3
pOH = - log [OH-] = - log 2 x 10-3 = 3 - log 2
pH = 14 - pOH = 11 + log 2