# What is the pOH of a 0.001 M HCl solution?

May 15, 2017

Well, we know that $p H + p O H = 14$, so..............

$p O H = 11$

#### Explanation:

For the autoprotolysis reaction:

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

we know that ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions............

And taking ${\log}_{10}$ of both sides we gets............

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

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

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

So................

$p H + p O H = 14$

You will not be asked to reproduce this as a first year student or as an A-level student. You will be asked (as you have been here!) to use the relationship.

And $\left[H {O}^{-}\right] = {10}^{- 11} \cdot m o l \cdot {L}^{-} 1$.