# A solution contains [OH^-] = 4.0 times 10^-5 M, what is the concentration of [H_3O^+]?

May 25, 2017

$\left[{H}_{3} {O}^{+}\right] = 2.50 \times {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........

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

Under standard conditions............

${K}_{\text{rxn}} = {K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$. And typically we would use logarithms to reduce this expression to.......

$p H + p O H = 14$, where the $p H$ function means $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ etc.

So $p O H = - {\log}_{10} \left[4.0 \times {10}^{-} 5\right] = 4.40$, and $p H = 9.60$.

And thus $\left[{H}_{3} {O}^{+}\right] = {10}^{- 9.60} = 2.50 \times {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1.$

May 25, 2017

$\left[{H}_{3} {O}^{+}\right] = 2.5 \cdot {10}^{- 10} M$

#### Explanation:

Given that no hydrolysis happened in the solution (inferred from question) Calculate $\left[{H}_{3} {O}^{+}\right]$ from ${k}_{w}$ directly.
$\left[{H}_{3} {O}^{+}\right]$
$= {k}_{w} / \left(\left[O {H}^{-}\right]\right)$
$= {10}^{-} \frac{14}{4.0 \cdot {10}^{-} 5}$
$= 2.5 \cdot {10}^{-} 10 M$

May 25, 2017

$2.5 \times {10}^{-} 10$ $\text{M}$ 