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Question
Derive the equation pH + pOH = 14.
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Solution
The ionic product of water is given as:
Kw = [H3O+][OH–]
Now, Kw = 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
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