Jika harga kb NH3=2.10^-5 tentukan pH larutan NH3 0,2 molar

[OH-] = √(Kb x M)
[OH-] = √(2 x 10^-5 x 2 x 10^-1)
[OH-] = √(4 x 10^-6)
[OH-] = 2 x 10^-3
pOH = 3 - log2

pH = 14 - pOH
pH = 14 - (3 - log2)
pH = 11 + log2