# How do you calculate the pOH of a solution?

Jun 6, 2016

In water you use the relationship $p H + p O H = 14$.

#### Explanation:

We know that at $298 \cdot K$, equilibrium obtains between hydronium ion and hydroxide ions in water:

${K}_{w} = {10}^{- 14} = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]$. We can take $- {\log}_{10}$ of both sides:

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

i.e. $- {\log}_{10} \left[H {O}^{-}\right] - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 14$ by definition of the log function. And again by definition $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ etc.

Thus $p H + p O H = 14$. There should be many problems here that use this relationship. Of course, you have to find them. Good luck.