# What is the pH of a 0.1 M acid solution?

Jun 9, 2016

For a strong acid, $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} \left[0.1\right] = - {\log}_{10} \left({10}^{-} 1\right)$ $=$ $- \left(- 1\right)$ $=$ $1$

#### Explanation:

We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.

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

Why do we use $p H$? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).

When I write ${\log}_{a} b = c$, I explicitly say that ${a}^{c} = b$. Common logarithmic bases are $10$ and $e$ (I think you cover this in A level mathematics these days). $p H$ and $p {K}_{a}$ scales use logarithms to the base $10$.

Anyway given what I have said ${\log}_{10} 100 = 2$, ${\log}_{10} 1000 = 3$, and ${\log}_{10} 1 = 0$, and ${\log}_{10} 0.1 = {\log}_{10} {10}^{-} 1 = - 1$. If you can grasp this, you will find the $p H$ concept (literally $\text{pouvoir hydrogene, power of hydrogen}$) fairly straightforward.

With weaker acids, say acetic acid, ${H}_{3} C - C \left(= O\right) O H$, the acid-base equilibrium lies to the left. The equilibrium constant, ${K}_{a}$, can be measured and often reported as $p {K}_{a}$ $=$ $- {\log}_{10} {K}_{a}$.