# What is the relationship between the pOH and the OH- ion concentration of a solution?

Dec 16, 2016

$p O H = - {\log}_{10} \left[H {O}^{-}\right]$

#### Explanation:

And in aqueous solution, the acid base equilibrium operates:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

At $298 \cdot K$, ${K}_{w} = {10}^{-} 14 = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]$

And taking negative ${\log}_{10}$ of both sides.............

$- {\log}_{10} {K}_{w} = - {\log}_{10} {10}^{-} 14 = - {\log}_{10} \left[H {O}^{-}\right] - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

i.e. $14 = p H + p O H$.