# Why is dissociation of water important?

Dec 31, 2017

Well, it underlies our understanding of acid-base behaviour....

#### Explanation:

Water is just another solvent, admittedly it is a very important solvent, and one that provides the basis for ALL of biochemistry.

In the bulk solvent we conceive of acids, characteristic cations, and base, characteristic anions:

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

Alternatively.....

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

Just to note that in aqueous solution under standard conditions, the ion product...

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

And we can take ${\log}_{10}$ of both sides to give....

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

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

Or.....

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

$14 = p H + p O H$

This relationship allows us to make very accurate assessments of $p H$ and acidity with relatively simple apparatus. Even with A level equipment, you can get exceptionally accurate and meaningful results in acid-base titrations.