Are pH and pOH related to each other?

May 7, 2016

Absolutely! In water:

$p H + p O H = 14$

Explanation:

We know that water undergoes autoprotolysis:

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

This is an equilibrium reaction, measured precisely over a range of temperatures. It is a fact that at $298 K$,

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${10}^{- 14}$ ($\left[{H}_{2} O\right]$ is treated as a constant!). Now this is a mathematical equation, which I can manipulate, as long as I do the same thing to both sides.

Taking ${\log}_{10}$ of both sides:

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

On rearrangement,

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

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and ${\log}_{10} \left[H {O}^{-}\right] = p O H$.

So $p H + p O H = 14$