sebanyak 100 ml larutan NH3 0,001 M (kb=10^-5) memiliki pH sebesar

OH- = √kb x Mb
= √10^-5 x 10^-3
= √ 10^-8
= 10^4
pOH = - log OH-
= - log 10^-4
= 4
pH = pKw - pOH
= 14 - 4
= 10

Larutan
Kimia XI

Basa lemah NH₃

[OH⁻] = √ [Kb x M]

[OH⁻] = √ [10⁻⁵ x 10⁻³]

[OH⁻] = 10⁻⁴

pOH = 4

Jadi pH = 14 - 4 = 10