# How do we get "pH" from "pOH"?

Jan 10, 2017

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

#### Explanation:

In aqueous solution we use the relationship, $p H + p O H = 14$, where $p H = - {\log}_{10} \left[{H}^{+}\right]$.

So if $p O H$ is high, then $p H$ is on the low end, i.e. an acidic solution with $p H$ around zero, and substantial $\left[{H}^{+}\right]$; and thus last option, low $p H$ and high $\left[{H}^{+}\right]$ or $\left[{H}_{3} {O}^{+}\right]$.