How do you find pOH given pH?

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Answer 1 · 2016-05-09T21:13:08

$p O H = 14 - p H$ $pOH=14-pH$

To find the $p O H$ $pOH$ from the given $p H$ $pH$ you can simply use the following relationship:

$p O H = 14 - p H$ $pOH=14-pH$

This relationship is derived from the expression of the water dissociation constant $K_{w}$ $K_w$ , where,

$K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$ $K_w=[H^(+)][OH^(-)]=1.0xx10^(-14)$

Taking the $- log$ $-log$ of both parties we get:

$- {log K}_{w} = - log [H^{+}] - log [O H^{-}] = 14$ $-logK_w=-log[H^(+)]-log[OH^(-)]=14$

Therefore, $p H + p O H = 14 \Rightarrow p O H = 14 - p H$ $pH+pOH=14=>pOH=14-pH$

remember, $p H = - log [H^{+}]$ $pH=-log[H^(+)]$ and $p O H = - log [O H^{-}]$ $pOH=-log[OH^(-)]$

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Acids & Bases | Nature, Strength & pH Scale.