# What is [HO^-] in an aqueous solution for which [H_3O^+]=3.2xx10^-3*mol*L^-1?

Jul 30, 2017

#### Answer:

$\left[H {O}^{-}\right] = 3.16 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We know that water undergoes autoprotolysis according to the following equation......

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

WE know by precise measurement at $298 \cdot K$ under standard conditions, the ion product......

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

Now this is an equation, that I may divide, subtract, multiply, PROVIDED that I does it to both sides. One thing I can do is to take ${\log}_{10}$ of BOTH SIDES........

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

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

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

And so (FINALLY) we gets our working relationship, which you should commit to memory.....

$p H + p O H = 14$

We have $\left[{H}_{3} {O}^{+}\right] = 3.2 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$. $p H = - {\log}_{10} \left(3.2 \times {10}^{-} 3\right) = 2.50$. And thus $p O H = 11.50$.

And now we take antilogs.....$\left[H {O}^{-}\right] = {10}^{- 11.50} \cdot m o l \cdot {L}^{-} 1$

$= 3.16 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$ as required............