# What is pOH used for?

Jun 2, 2016

Both $p H$ and $p O H$ are used to simplify calculations.

#### Explanation:

As you know, in aqueous solution at $298 \cdot K$, $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, $p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$, AND $p H + p O H = 14$. Back in the day, before the advent of electronic calculators, students and engineers would routinely use log tables to enable complex multiplication and division. If I have a basic solution, I can thus calculate $\left[{H}_{3} {O}^{+}\right]$ on the basis of $p O H$ by means of simple addition or subtraction rather than by more complex multiplication/division.

If you are still unsure, come back here, and someone will help you.

In the meantime, ${K}_{w}$ =([H_3O^+][HO^-])=10^(14)) under standard conditions, i.e. $298 \cdot K$, and $1 \cdot a t m$. If we were to increase the temperature, say to water's normal boiling point, $373 \cdot K$, how would you predict ${K}_{w}$ and $p {K}_{w}$ to evolve?

Remember that the autoprotolysis of water is an example of a bond-breaking reaction.

$2 {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + H {O}^{-}$