# Why are pH and pOH defined on the logarithmic scale?

Feb 20, 2017

Let's look at the $\text{logarithmic function}$ first............

#### Explanation:

When we use logarithms, we usually have a function of the form ${\log}_{a} b = c$. ${\log}_{a} b$ asks for what power I raise the base $a$ to get $b$. Here, since ${\log}_{a} b = c$, then BY DEFINITION, ${a}^{c} = b$.

To put this in a bit more practical way, we know that ${10}^{2} = 100$, and ${10}^{3} = 1000$. If we take logarithms to the base 10, then ${\log}_{10} 100 = 2$, and ${\log}_{10} 1000 = 3$. Why? Because ${10}^{2} = 100$, and ${10}^{3} = 1000$. And when I input a number on my calculator, say $x$, when I push the $\log$ button, I get a number $y$, such that ${10}^{y} = x$. Try this out with $100$, $1000$, $1000000$. You will get simple whole numbers.

Back in the day, before the advent of electronic calculators, students of maths and chemistry and physics, would routinely use logarithmic tables for multiplying or dividing large (or very small) numbers.

In water, it is a fact that for the ion product given in the equation,

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

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ at $298 K$. I can take $\text{logarithms to the base 10}$ of BOTH sides to give:

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

But CLEARLY, the power to which I raise $10$ to get ${10}^{-} 14$ is $- 14$. Are you with me?

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

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

And if I DEFINE $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, then we get the defining relationship:

$p H + p O H = 14 \text{ in water at 298K}$.

More on buffers here if you need it.