# Is a solution with a pOH of 12.1 acidic, basic, or neither?

Apr 25, 2017

This value specifies a highly acidic solution...............$p H = 1.9$

#### Explanation:

For the acid-base equilibrium that operates in water.....

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

We know that at under standard conditions, $298 \cdot K , \text{1 atmosphere}$.............

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

This defines the acid-base equilibrium, the autoprotolysis of water.

We can take ${\log}_{10}$ of BOTH sides........

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

And on rearrangement,

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

Given that ${K}_{w} = {10}^{-} 14$, and $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, then BY DEFINITION, ${\underbrace{- {\log}_{10} {K}_{w}}}_{p {K}_{w}} = - {\log}_{10} {10}^{- 14} = - \left(- 14\right) = 14$, and the defining relationship, which you may not have to derive, but WILL have to remember,

$14 = p H + p O H$

And given that $p O H = 12.1$, this means that $\left[H {O}^{-}\right] = {10}^{- 12.1} \cdot m o l \cdot {L}^{-} 1$, and $p H = 14 - 12.1 = 1.9$, and (FINALLY) $\left[{H}_{3} {O}^{+}\right] = {10}^{- 1.9} \cdot m o l \cdot {L}^{-} 1 = 0.0126 \cdot m o l \cdot {L}^{-} 1.$

Now this might seem a lot of work, but only because I derived the equation. You must be able to use the relationship...........

$14 = p H + p O H$

These logarithmic terms were introduced back in the day, before the advent of electronic calculators. Log tables, printed values of ${\log}_{10}$ and ${\log}_{e}$ were widely used by scientists, engineers, and by students of course.