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Derive the relationship between pH and pOH.

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Question

Derive the relationship between pH and pOH.

Answer in Brief
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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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2021-2022 (March) Set 1

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