Given the pH of the solution, we can determine [H⁺].
pH = -log[H⁺]
[H⁺] = 10⁻⁷·⁷⁸
[H⁺] = 1.659586907 x 10⁻⁸ M
We also know that pOH = 14 - pH. Thus, the pOH is 6.22. Given the pOH of the solution, we can determine [OH⁻].
pOH = -log[OH⁻]
[OH⁻] = 10⁻6.22
[OH⁻] = 6.02559586 x 10⁻⁷ M
The ionic product of water Kw, can then be calculated using the following equation.
Kw = [H⁺][OH⁻]
Kw = (1.659586907 x 10⁻⁸)(6.02559586 x 10⁻⁷)
Kw = 1.0 x 10⁻¹⁴
