Question

Derive the equation p^H + p^OH = 14.

Answer 1

The ionic product of water is given as:

K_w = [H₃O⁺][OH^–]

Now, K_w = 1 × 10^–14 at 298 K

Thus, [H₃O⁺][OH^–] = 1.0 × 10^–14

Taking logarithm of both the sides, we write

log₁₀[H₃O⁺] + log₁₀[OH^–] = –14

–log₁₀[H₃O⁺] + {– log₁₀[OH^–]} = 14

Now, pH = –log₁₀[H₃O⁺] and pOH = –log₁₀[OH^–]

∴ pH + pOH = 14