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[OH-] = [KOH] x 1
[OH-] = 4 x 10^-3
pOH = -log [OH-] = -log 4x10^-3
pOH = 3 - log 4 = 3 - 0,6 = 2,4
pH = 14 - pOH = 14 - (3 - log 4)
pH = 14 - 3 + log 4 = 11 + log 4 = 11,6
* pers reaksi KOH = K^+ + OH^-, jadi di dapat n = 1
* [OH^-] = n x M = 1 x 4 x 10^-3 = 4 X 10^-3
* pOH = - log [OH^-] = - log 4 x 10^-3 = 3 - log 4 = 3 - 0,6 = 2,4
* pH = 14 - 2,4 = 11,6
jadi pHnya = 11,6