How do you convert pH to pOH?

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Answer 1 · 2016-05-16T14:33:29

$pOH = 14.00 - pH$ $"pOH" = "14.00 - pH"$

$H^{+} + O H^{-}$ $H^+ + OH^-$ $⇌$ $rightleftharpoons$ $H_{2} O$ $H_2O$

the ionic product for water, $K_{w}$ $K_w$ = $10^{- 14}$ $10^-14$

$\Rightarrow K_{w} = ([O H^{-}] [H^{+}])$ $=> K_w = ([OH^-][H^+])$

applying log on both sides,

-14= $log [O H^{-}] + log [H +]$ $log[OH^-] +log[H+]$

14= $- log [O H^{-}] - log [H +]$ $-log[OH^-] -log[H+]$

therefore 14= pH +pOH

(since $p H = - log [H^{+}])$ $pH= -log[H^+])$

$[H^{+}]$ $[H^+]$ = concentration of $H^{+}$ $H^+$