# How do we define pH?

Aug 25, 2017

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

#### Explanation:

$p H$ is a measure of hydrogen ion concentration in aqueous solution. In water, we know that it undergoes the so-called $\text{autoprotolysis reaction}$:

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

And at $298 \cdot K$, we know that ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$.

And this is an equation, that we can divide, multiply, add to, subtract from....etc. PROVIDED that we do it to both sides of the equality.

One thing we can do is to take ${\log}_{10}$ of both sides....

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

On rearrangement......

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

But $- {\log}_{10} {K}_{w} = - {\log}_{10} \left({10}^{-} 14\right) = - \left(- 14\right)$, and thus, finally,

$14 = p H + p O H$ in aqueous solution under standard conditions. How do you think ${K}_{w}$ would evolve under NON-STANDARD conditions, i.e. $T = 100$ ""^@C. Remember that the autoprotolysis reaction is bond-breaking.