# If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid?

Jul 28, 2016

$p {K}_{a} = 6.52$

#### Explanation:

If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:

$H A \text{ "+H_2O""rightleftharpoons""H_3O^+""+" } {A}^{-}$

$\textcolor{red}{I} \text{ "1" "mol" " 0" "mol" " 0" " } m o l$

$\textcolor{red}{C} \text{ "-alpha" "mol" " alpha" "mol" " alpha" " } m o l$

$\textcolor{red}{E} \text{ "1-alpha" "mol" " alpha" "mol" " alpha" " } m o l$

Where $\alpha$ is the degree of dissociation of $H A$

If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows

$\left[H A\right] = c \left(1 - \alpha\right) M$

$\left[{H}_{3} {O}^{+}\right] = c \alpha M$

$\left[{A}^{-}\right] = c \alpha M$

The concentration of ${H}_{2} O$ is irrelevant as it acts as sovent.

Now the acid dissociation constant

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{H A} ^ 2$

$= \frac{\left(c \alpha\right) \cdot \left(c \alpha\right)}{c \left(1 - \alpha\right)}$

$= \frac{c {\alpha}^{2}}{1 - \alpha}$

Given c=6.03xx10^-6M and alpha=20%=0.2

${K}_{a} = \frac{6.03 \times {10}^{-} 6 \cdot {\left(0.2\right)}^{2}}{1 - 0.2}$

$= \frac{6.03 \times {10}^{-} 6 \times 0.04}{0.8}$

$3.015 \times {10}^{-} 7$

The $p {K}_{a}$ of the weak acid

$p {K}_{a} = - \log {K}_{a} = - \log \left(3.015 \times {10}^{-} 7\right)$

$= 7 - \log 3.015 = 6.52$