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Question
Derive the relationship between pH and pOH.
Solution
The ionic product of water is
KW = \[\ce{[H3O+][OH–]}\]
Now, KW = 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
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