Question
Derive the equation $p^H + p^{OH} = 14.$

Answer

The ionic product of water is given as:
$K_w = [H_3O^+][OH^–]$
Now, $K_w = 1 \times 10^{–14} at 298 K$
Thus, $[H_3O^+][OH^–] = 1.0 \times 10^{–14}$
Taking logarithm of both the sides, we write
$log_{10}[H_3O^+] + log_{10}[OH^–] = –14$
$–log_{10}[H_3O^+] + {– log_{10}[OH^–]} = 14$
Now, p$H = –log_{10}[H_3O^+]$ and $pOH = –log_{10}[OH^–]$
$\therefore pH + pOH = 14$

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