Question

Derive the relationship between pH and pOH.

Answer 1

The ionic product of water is

K_W = \[\ce{[H3O+][OH–]}\]

Now, K_W = 1 × 10⁻¹⁴ at 298 K and thus

\[\ce{[H3O+][OH–]}\] = 1 × 10⁻¹⁴

Taking logarithm of both sides,

\[\ce{log10[H3O+] + log10[OH–]}\] = −14

Multiply the entire equation by '−1'

\[\ce{(-log10[H3O+]) + (-log10[OH–])}\] = 14

Now pH = \[\ce{- log10[H3O+]}\] and pOH = \[\ce{- log10[OH–]}\]

∴ pH + pOH = 14