Larutan Ca(OH)2 0,01M sebanyak 1L memiliki pH sebesar

[OH-]=M.Valensi
=0.01X2
=0.02=2×10^-2
pOH=log[OH-]
=2-log2
=2-0.3
=1.7
pH=14-1.7=12.3

Larutan
Kimia XI

Ca(OH)₂ sebagai basa kuat bervalensi 2

[OH⁻] = valensi x M

[OH⁻] = 2 x 10⁻²

pOH = - log [OH⁻]

pOH = - log [2 x 10⁻²]

pOH = 2 - log 2

pH = 14 - pOH

pH = 14 - (2 - log 2)

Jadi pH = 12 + log 2