# A vinegar bottle has a hydroxide concentration of 5.0*10^-12 M. What is the hydronium concentration?

Aug 30, 2017

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 2.70} \cdot m o l \cdot {L}^{-} 1 = 2.0 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We know that water undergoes autoprotolysis, and that this reaction can be precisely quantified, i.e.

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

This is an equilibrium reaction, and under standard conditions,

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ (this is usually treated as a dimensionless quantity, and so.........

....when we take ${\log}_{10}$ of BOTH sides, we get......)

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

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

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

And thus we get the working relationship.....

$p H + p O H = 14$ IN AQUEOUS SOLUTION UNDER STANDARD CONDITIONS. And this is something you will habitually use in acid-base equilibria.

So we gots an ACIDIC solution, where $\left[H {O}^{-}\right] = 5.0 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$, and thus $p O H = - {\log}_{10} \left(5.0 \times {10}^{-} 12\right) = - \left(11.30\right) = 11.3$.

And $p H = 14 - p O H = 14 - 11.3 = 2.7$ And thus (FINALLY) [H_3O^+]=10^(-pH)=....?

I acknowledge that I have included a lot of background in this question. But if you can follow it, and use logarithms effectively, these sorts of questions become quite trivial. If there is an issue for clarification, please report it.