It has been found that the pH of a 0.01M solution of an organic acid is 4.15.
Question:

It has been found that the $\mathrm{pH}$ of a $0.01 \mathrm{M}$ solution of an organic acid is $4.15$. Calculate the concentration of the anion, the ionization constant of the acid and its $\mathrm{pK}_{\mathrm{a}}$.

Solution:

Let the organic acid be HA.

$\Rightarrow \mathrm{HA} \longleftrightarrow \mathrm{H}^{+}+\mathrm{A}^{-}$

Concentration of $\mathrm{HA}=0.01 \mathrm{M}$

$\mathrm{pH}=4.15$

$-\log \left[\mathrm{H}^{+}\right]=4.15$

$\left[\mathrm{H}^{+}\right]=7.08 \times 10^{-5}$

Now, $K_{a}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$

$\left[\mathrm{H}^{+}\right]=\left[\mathrm{A}^{-}\right]=7.08 \times 10^{-5}$

$[\mathrm{HA}]=0.01$

Then,

$K_{o}=\frac{\left(7.08 \times 10^{-5}\right)\left(7.08 \times 10^{-5}\right)}{0.01}$

$K_{a}=5.01 \times 10^{-7}$

$p K_{a}=-\log K_{a}$

$=-\log \left(5.01 \times 10^{-7}\right)$

$p K_{a}=6.3001$