# What is pH and how is it defined?

Apr 20, 2017

$p H$ $=$ $\text{pouvoir hydrogene}$

#### Explanation:

So $p H$ literally stands for $\text{power of hydrogen}$, and was introduced in the early 20th century by the Danish chemist, Soren Sorenson.

Normally, we use it to measure the concentration of hydrogen ion in aqueous solution............

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

$\left[{H}_{3} {O}^{+}\right]$ is a conceptual species, and could also be represented as ${H}^{+}$. It is the characteristic cation of the water solvent; and the characteristic anion of water is designated as hydroxide, ""^(-)OH.

By careful measurement of pure water, $\left[{H}_{3} {O}^{+}\right]$ and $\left[H {O}^{-}\right]$, were found to be present in some (small) quantity at equilibrium dependent on temperature:

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

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ at $298 \cdot K$ (NB $\left[{H}_{2} O\right]$ does not appear in the equilibrium expression, because it is so large it is effectively constant.) So in an acid solution, the relationship tells us that [""^(-)OH] is low, and vice versa for an alkaline solution.

Now back in the day (only 30-40 years ago in fact), before the advent of cheap electronic calculators, division and multiplication of large and small numbers was fairly difficult. Scientists, students and teachers, and engineers would routinely use logarithmic tables (to the $\text{base 10}$ or $\text{base e}$ for multiplication and division.)

And thus the product axxb-=10^("log"_10 a +"log"_10 b". It was easier to find $\text{logs}$ and $\text{antilogs}$ of the given product, than it was to do the long multiplication/division. Of course these days we would simply plug the numbers into a hand-held calculator.

But if we take ${\log}_{10}$ of ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$, we gets.....

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

But since ${\log}_{a} {a}^{b} = b$ by definition, then, ${\log}_{10} {10}^{-} 14 = - 14$.

And so -14=log_10[H_3O^+]+log_10[""^(-)OH]

OR

14=-log_10[H_3O^+]-log_10[""^(-)OH]

And given our definition,

$14 = p H + p O H$

And thus a low or negative $p H$ characterizes a STRONGLY acidic solution, and a high $p H$ (approaching $14$) characterizes a STRONGLY alkaline solution.

I have gone on for a bit about nothing to the power of less. Look at the examples of $p H$ in your text to consolidate your understanding.