ix. pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant

\begin{align*} \text{Given:} \\ \text{pH} &= 3.2 \\ \text{Concentration Dissociation} &= 0.02 \, \text{M} \\ \text{Dissociation constant (} K_a \text{)} &= ? \\ \text{Calculation:} \\ K_a &= \frac{\left(\alpha^2\right)}{(1 - \alpha)} \\ \alpha &= \frac{[H^+]}{C} \\ \text{pH} &= -\log_{10}[H^+] \\ 3.2 &= -\log_{10}[H^+] \\ -3.2 &= \log_{10}[H^+] \\ 10^{-3.2} &= [H^+] \\ 6.3 \times 10^{-4} &= [H^+] \\ \alpha &= \frac{[H^+]}{C} \\ &= \frac{6.3 \times 10^{-4}}{0.02} \\ &= 0.0315 \\ K_a &= \frac{\left(0.02 \times 0.0315\right)^2}{1 - 0.0315} \\ &= \frac{0.02 \times 0.000992}{0.9685} \\ &= \frac{0.02 \times 9.92 \times 10^{-4}}{1} \\ &= 0.1984 \times 10^{-4} \end{align*}