Hitunglah ph dari 150ml larutan NH4Cl 0.008M, jika diketahui Kb NH4OH 1x10^-5!

pH = 1/2 (14 - pKb - log[G])
pH = 1/2 (14 - (-log10^-5) - log 8 x 10^-3)
pH = 1/2 (14 - 5 + 3 - log8)
pH = 1/2 (12 - log8)
pH = 6 - log8/2
pH = 6 - 0,45
pH = 5,55

→ Atau :

[H+] = √Kw / Kb . M
[H+] = √10^-14 / 10^-5 . 8 x 10^-3
[H+] = √10^-9 x 8 x 10^-3
[H+] = √8 x 10^-6

pH = -log [H+]
pH = -log√8 x 10^-6
pH = 6 - log√8
pH = 6 - 0,45
pH = 5,55