# How do pH values of acids and bases differ?

Jun 20, 2018

We assess the autoprotolysis of water...

#### Explanation:

...which reaction we represent as...

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

Now this reaction is governed by an equilibrium constant....and we write it in the usual way...and of course the EXTENT of equilibrium can be measured, and so we can quantify it...

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

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

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

And so on rearrangement....

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]}}_{\text{pH+pOH by definition}} = + 14$

And thus our defining relation...$p H + p O H = 14$ for water under standard conditions.... And in stronger acids...$p H$ tends low to negative, i.e. $\left[{H}_{3} {O}^{+}\right]$ is high...and in bases $p O H$ tends low to negative...and $p H$ tends towards $14$.

Using the same sort of approach, for conjugate acid/conjugate base pairs...$p {K}_{a} + p {K}_{b} = 14$...in aqueous solution...