# If [H_3O^+]=1.2xx10^-8*mol*L^-1, what are pH and pOH?

Apr 18, 2017

We know that $p H + p O H = 14$ under standard conditions in an aqueous medium.

#### Explanation:

So some (necessary?) background. Water undergoes autoprotolysis according to:

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

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

As with any equation, we can divide/multiple/add/substract etc. PROVIDED that we do it to both sides of the equation. One thing we can do (and remember that this was important in the days before electronic calculators!) is to take ${\log}_{10}$ of both sides..........

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

But ${\log}_{a} {a}^{b} = b$ by definition.............and so ${\log}_{10} {10}^{-} 14 = - 14$.

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

OR...............multliplying each side by $- 1$

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

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

$p H + p O H = 14$

And you have to know this derivation as a 1st year undergraduate but not as an A-level student........

And now it is simply a matter of taking your trusty electronic calculator, and pushing the $\text{log button}$.

[H_3O^+]=1.2xx10^-8; pH=-log_(10)1.2xx10^-8=-(-7.92)=7.92

$p O H = 14 - 7.92 = 6.08$

And now we take antilogs..........

$\left[H {O}^{-}\right] = {10}^{-} 6.08 \cdot m o l \cdot {L}^{-} 1 = 8.33 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1.$

I leave it to you to answer the 2nd problem, which I hope you will post here.