# What is the [OH^-] for a solution at 25°C that has a [H_3O^+] of 5.36 times 10^-5?

Jun 22, 2017

$\left[H {O}^{-}\right] = 1.87 \times {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

Given.......

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

${K}_{w} ' = \frac{\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{2} O \left(l\right)\right]}$

Because $\left[{H}_{2} O\right]$ is so large, we remove it from the expression to give,,,,,,

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions of temperature $\left(298 \cdot K\right)$ and pressure $\left(1 \cdot a t m\right)$...

Thus $\left[H {O}^{-}\right] = {10}^{-} \frac{14}{5.36 \times {10}^{-} 5} = 1.87 \times {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1$.

Alternatively, we could take ${\log}_{10}$ of both sides of ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ to give.........

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

On rearrangement

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\textcolor{red}{p H}} {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{\textcolor{b l u e}{p O H}} = {\underbrace{- {\log}_{10} {K}_{w}}}_{p {K}_{w}}$

And thus $\textcolor{red}{p H} + \textcolor{b l u e}{p O H} = p {K}_{w} = 14$.

And here, $p H = - {\log}_{10} \left(5.36 \times {10}^{-} 5\right) = 4.27$, and clearly, $p O H = 9.73$.

$\left[H {O}^{-}\right] = {10}^{- 9.73} = 1.87 \times {10}^{-} 10 \cdot m o l \cdot {L}^{-} 1$, as required.......

I don't know about you, but I think the $\log$ method is better.....