Suatu basa organik memiliki 0.1 M larutan. PH =11. tentukan harga Kb
rikarahma123
PH = 11 pOH = 14 - 11 = 3 - log 1 [OH-] = 1 . 10^-3 [OH-] = √Kb . Mb [1.10^-3]² =( √Kb . 0,1)² 1.10^-6 = Kb . 0,1 Kb = 1.10^-6 / 1.10^-1 = 1.10^-5 Semoga membantu
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