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n NH4Cl = 500 mL x 0,1 M = 50 mmol
[OH-] = Kb x n NH3 ÷ n NH4Cl
= 1,8 x 10^-5 x 10 mmol ÷ 50 mmol
= 3,6 x 10^-6
pOH= -log 3,6 x 10^-6
= 6 - log 3,6
= 6 - 0,556 = 5,444
pH = 14 - 5,444 = 8,556 atau
= 14 - (6 - log 3,6) = 8 + log 3,6