# What is the logarithmic concentration of the hydroxide ion in aqueous solution?

Jan 26, 2016

I presume you mean $p O H$?

#### Explanation:

We would calculate $p O H$ the same way we calculate $p H$. That is $p O H$ $=$ $- {\log}_{10} \left[O {H}^{-}\right]$, where $\left[O {H}^{-}\right]$ is the concentration of hydroxide ion.

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

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

And thus $14$ $=$ $p H + p O H$ (because by definition ${\log}_{10} {10}^{- 14}$ $=$ $- 14$.

So you need to find hydroxide (or acid) concentration, and take its negative logarithm to the base 10.