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Air mempunyai kesetimbangan seperti berikut:
2H₂O ⇔ H₃O⁺ + OH⁻
karena konsentrasi [H₂O] sangat kecil maka nilainya diabaikan, maka konstata kesetimbangannnya adalah
Kw = [H₃O⁺] [OH⁻] atau Kw = [H⁺] [OH⁻]
jika Kw diketahui 1,2 . 10 ⁻⁵, maka
Kw = [H⁺] [OH⁻]
1,2 . 10 ⁻⁵ = [H⁺] [OH⁻]
karena H₂O bersifat netral maka [H⁺] = [OH⁻] ; atau selanjutnya kita misalkan x M
12 . 10 ⁻⁶ = [H⁺] [OH⁻]
12 . 10 ⁻⁶ = X . X
12 . 10 ⁻⁶ = X²
X = √(12 . 10 ⁻⁶)
X = 3,46 . 10 ⁻³
Jadi [H⁺] = 3,46 . 10 ⁻³
Maka Ph = - log [H⁺]
= - log 3,46 . 10 ⁻³
= 3 - log 3,46