Di dalam 200 ml larutan terlarut 5,35 gram NH4CL jika Kb=10^-5 , Ar N=14 , H =1 ,CL = 35,5 hitunglag pH larutan tersebut

Mr = 14 + 4 + 35,5
Mr = 53,5

mol = gr / Mr
mol = 5,35 / 53,5
mol = 0,1 mol

M = mol / liter
M = 0,1 / 0,2
M = 0,5

Kb = tetapan ionisasi basa = 10^-5

[OH-] = √kb [NH4CL]
[OH-] = √10^-5 (0,5)
[OH-] = √5 x 10^-6
[OH-] = √5 x 10^-3

pOH = - log [OH-]
pOH = 3 - log√5

pH = 14 - (3 - log√5)
pH = 14 - 3 + log√5
pH = 11 + log√5