# If the pH=9 for a solution of KOH(aq), what is the concentration of HO^-?

## $A .$ $9 \cdot m o l \cdot {L}^{-} 1$; $B .$ $5 \cdot m o l \cdot {L}^{-} 1$; $C .$ ${10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$; $D .$ $\text{Cannot be determined.}$

Aug 18, 2016

$\text{Option C}$

#### Explanation:

It is a fact that in aqueous solution under standard condtions,

$p H + p O H$ $=$ $14$

If $p H$ $=$ $9$, then $p O H = 14 - 9 = 5$

And thus, given the definition of $p O H$, $\left[H {O}^{-}\right]$ $=$ ${10}^{- 5} \cdot m o l \cdot {L}^{-} 1$.

Just to expand this a little bit, we can go back to the autoprotolysis of water:

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

Now we can certainly write the equilibrium expression for this rxn:

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

At $298$ $K$ and $1 \cdot a t m$, ${K}_{w}$ has been precisely and accurately measured, and the $\left[{H}_{2} O\right]$ term can be removed because it is so large it is effectively constant. And thus,

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

Now this is a mathematical expression, which we can divide, mulitply, etc. PROVIDED that we do it to both sides of the equation. One thing that can be done (and was done routinely in the days before electronic calculators) was take ${\log}_{10}$ of both sides:

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

OR

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

But,
$- {\log}_{10} {10}^{-} 14 = - \left(- 14\right) = 14$, and $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and $- {\log}_{10} \left[H {O}^{-}\right]$ $=$ $p O H$. All of this is by definition.

And thus $p H + p O H = 14$.

This last is certainly worth committing to memory, or if you are an undergrad, commit the entire derivation to memory.