# Represent the autoprotolysis of water, and explain how pH defines the acidity of the solution. How are pH and pOH defined?

Sep 4, 2017

$5.$ $2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

Note that $\left[{H}_{3} {O}^{+}\right]$ is synonymous with $\left[{H}^{+}\right]$

#### Explanation:

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

And ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ at $298 \cdot K$. This equilibrium has been extensively measured.

$\text{6. and 7}$ We can take ${\log}_{10}$ of both sides...........

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

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

And so on rearrangement.........

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

But by definition........

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

And so $14 = p H + p O H$, under standard conditions of $298 \cdot K$ and near $1 \cdot a t m$.

If $p H = 7$, then, clearly, $\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right]$, and the solution is NEUTRAL.

If $p H < 7$, then, clearly, $\left[{H}_{3} {O}^{+}\right] > \left[H {O}^{-}\right]$, and the solution is ACIDIC.

If $p H > 7$, then, clearly, $\left[{H}_{3} {O}^{+}\right] < \left[H {O}^{-}\right]$, and the solution is ALKALINE.

$8.$ We gots $\left[{H}_{3} {O}^{+}\right] = 3.5 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$.

And given the former....$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(3.5 \times {10}^{-} 8\right)$

$= - \left(- 7.46\right) = 7.46$, i.e. a basic solution where $\left[{H}_{3} {O}^{+}\right]$.

$p O H = 14 - 7.46 = 6.54$, and $\left[H {O}^{-}\right] = {10}^{- 6.54} \cdot m o l \cdot {L}^{-} 1 = 2.85 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.

And if we gots $\left[{H}_{3} {O}^{+}\right] = 0.0065 \cdot m o l \cdot {L}^{-} 1$, $p H = - {\log}_{10} \left(0.0065\right) = - \left(- 2.19\right) = 2.19$.

I leave it to you to calculate $p H$ and $p O H$ given the defining relationship.....$14 = p H + p O H$