# How is pOH determined?

Jun 4, 2016

$p O H$ is detemined equivalently to $p H$, but $p H + p O H = 14$ under standard conditions.

#### Explanation:

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

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

Now ${K}_{w}$ $=$ $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${10}^{- 14} \text{ at 298K}$.

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

$= - {\log}_{10} \left({10}^{-} 14\right)$ $=$ $+ 14$

And thus $p O H + p H = 14$ under standard conditions.

You should review the logarithmic function if you have any problem with this treatment. You should review it anyway, because I might have made a mistake.