Aug 22, 2017

$\left[H {O}^{-}\right] = 1.78 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

In aqueous solution, water undergoes the famous autoprotolysis reaction:

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

And under standard condition of $298 \cdot K$, and near $1 \cdot a t m$...

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

Taking "log_10 of BOTH sides........

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

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

And thus our working relationship.......which you will have to commit to memory:

$p H + p O H = 14$

We have $p H = 3.25$, thus $p O H = 10.75$, and .........

$\left[H {O}^{-}\right] = {10}^{- 10.75} = 1.78 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$......

For the use of $p H$ in buffer equations, see here.