# Why does pH + pOH = 14?

May 17, 2016

Because water undergoes a measurable equilibrium at $298 \cdot K$.

#### Explanation:

We can write the equilibrium as follows:

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

And, ${K}_{w}$ $=$ $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$. This equilibrium has been carefully measured (by conductivity experiments) over a range of temperatures. $\left[{H}_{2} O\right]$ does not appear in the equilibrium because its concentration is effectively constant.

At $298 \cdot K$ we know that ${K}_{w} = {10}^{-} 14.$ How do we know this? By careful and repeated measurements of the conductivity of the water solvent.

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

Now this is an arithmetic expression, that we can divide, multiply, add to, subtract from, provided that we do it to both sides of the equation. We can take ${\log}_{10}$ of both sides to give:

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

But, by definition of logarithms, if ${a}^{b} = c$, ${\log}_{a} c = b$, then ${\log}_{10} \left({10}^{-} 14\right) = - 14$, and thus,

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

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

$p {K}_{w} = p H + p O H = 14$ as required.

Given that this is a bond-breaking reaction, how would you expect the equilibrium to evolve at temperatures greater than $298 K$? Would $p H$ increase or decrease?