What does pOH mean?

May 26, 2016

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

Explanation:

So $p O H$ is the basic counterpart of $p H$ $\left(= - {\log}_{10} \left[{H}_{3} {O}^{+}\right]\right)$.

This derives from the autoprotolysis reaction in water:

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

This equilibrium has been carefully measured, and at $298 K$ we write,

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

We can take ${\log}_{10}$ of both sides to give:

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

${\log}_{10} {K}_{w}$ $=$ ${\log}_{10} \left[H {O}^{-}\right] + {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- 14$ (because by definition, ${\log}_{a} \left({a}^{b}\right) = b$. On rearrangement:

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

And, again by definition,

$p H + p O H = 14$

The autoprotolysis of water is a bond breaking reaction. How would you predict ${K}_{w}$ to evolve at elevated temperature? What would this mean with respect to $p H$ and $p O H$?