# What does pH measure?

Jan 23, 2017

$p H$ is a quantitative measure of the concentration of hydronium ion, ${H}_{3} {O}^{+}$ in aqueous solution.

#### Explanation:

Water is known to undergo autoprotolysis according to the following reaction:

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

This equilibrium has been exhaustively studied and we write the equilibrium in the normal way,

${K}_{\text{eq}} = \frac{\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{2} O\right]}$

Because $\left[{H}_{2} O\right]$ is LARGE, it can be assumed to be constant, and removed from the expression to give:

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ at $298 \cdot K$

A temperature is specified, because the extent of reaction depends upon temperature, especially for a bond-breaking reaction. Now this is a mathematical expression, the which we can divide, multiply etc., provided that we do it to BOTH sides of the expression. One think we can do is to take ${\log}_{10}$ of both sides for reasons that will become apparent later:

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

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

But ${\log}_{10} \left\{{10}^{-} 14\right\} = - 14$ by definition, and we can rearrange the given expression to give:

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

Of course, by definition $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$ and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$

So for water at $298 K$, $p H + p O H = 14$. This is the defining expression for acid base behaviour in water, and it is one with which you will get very familiar.

So to answer your question (finally!), $p H$ is a quantitative measure of the concentration of ${H}_{3} {O}^{+}$ (in water).

I apologize for going on so long, but you will need the given background if you don't know it already.