# How do you find the value of [OH^-] for a solution with a pH of 8.00?

Jun 14, 2017

$\left[H {O}^{-}\right] = {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$. How does we know.........?

#### Explanation:

Water undergoes so-called $\text{autoprotolysis}$......

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

At $298 \cdot K$ we can measure the $\text{autoprotolysis}$ very precisely....

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

This is a mathematical equation which we can divide, multiply, or otherwise manipulate provided that we does it to both sides.....One think that we can do is take ${\log}_{10}$ of both sides....

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

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

(Why? Because mathematically, ${\log}_{a} {a}^{y} = y$ by definition. It is the power to which we raise the $\text{base a}$ to get ${a}^{y}$. With me?

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

BUT, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$ BY DEFINITION........

And so (at last!) $14 = p H + p O H$

And so (after all that!) if $p H = 8$, then CLEARLY, $p O H = 6$. And thus $\left[H {O}^{-}\right] = {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$.

Claro?

Note the logarithmic function is something that you learn about in mathematics; it is very powerful, and useful. Its application to chemistry is (I think) rather straightforward. If there is still an issue, voice it and someone will help you.