# What is the pOH of a solution with a pH of .7?

Dec 11, 2016

And it's an aqueous solution?

#### Explanation:

We know that in water the following equilibrium operates:

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$ ${K}_{\text{eq}} = {10}^{-} 14$

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

We take ${\log}_{10}$ of both sides:

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

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = - 14$, and multiply both sides by $- 1$,

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

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

And thus $p H + p O H = 14$ (for water under standard conditions). This is something that you have to know.

You have $p H = 0.7$, and thus $p O H = 13.3$. What is $\left[{H}_{3} {O}^{+}\right]$?