# If [H_3O^+]=5.0xx10^-3*mol*L^-1, what is pH and what is pOH of this solution?

Apr 17, 2017

$p H = 2.30$............

#### Explanation:

.......and $p O H = 11.70$

These acid base reactions are governed by the equilibrium:

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

And at $298 \cdot K$, ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$

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

And since we know that ${K}_{w} = {10}^{-} 14$, then..........

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

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

But, by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p O H$

And thus, BY DEFINITON, $p H + p O H = 14$ under standard conditions. If you can remember this relationship, then these sorts of problems become almost trivial.

And thus, for your problem, $p H = - {\log}_{10} \left(0.0050\right) = - \left(- 2.30\right) = 2.30$.

$p O H = 14 - 2.30 = 11.70$.

And can you tell me what is $\left[H {O}^{-}\right]$ here? It should be small......