Tentukan ph larutan amoniak nh4oh 0.4 m, bila harga kb amoniak adalah 10^-5

[tex] {oh}^{ - } = \sqrt{kb \times \: m} \\ {oh}^{ - } = \sqrt{ {10}^{ - 5} \times 0.4 } \\ {oh}^{ - } = \sqrt{4 \times {10}^{ - 6} } \\ {oh}^{ - } = 2 \times {10}^{ - 3} \\ poh = 3 - log2 \\ ph = 11 + log2[/tex]