Derive the relation between $pH$ and $pOH$.

(N/A) The ionic equilibrium of dissociation of water at $298 \ K$ is given by:
$H_{2}O_{(l)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + OH^{-}_{(aq)}$
The ionic product of water is $K_{w} = [H_{3}O^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
Taking the negative logarithm on both sides:
$-\log K_{w} = -\log ([H_{3}O^{+}][OH^{-}]) = -\log (1.0 \times 10^{-14})$
Using the property $\log(ab) = \log a + \log b$:
$-\log K_{w} = -\log [H_{3}O^{+}] - \log [OH^{-}] = 14.0$
Since $pH = -\log [H_{3}O^{+}]$ and $pOH = -\log [OH^{-}]$,and $pK_{w} = -\log K_{w}$:
$pK_{w} = pH + pOH = 14.0$
Thus,the relation is $pH + pOH = 14.0$ at $298 \ K$.