JEE Main 2023 — Ionic Equilibrium Question with Solution

To find the dissociation constant of acetic acid, we use the Henderson-Hasselbalch equation for the given system and conditions. The equation is as follows:

$\text{pH}={\text{pK}}_{\text{a}}+\log \left(\frac{[{\text{CH}}_3{\text{COO}}^{-}]}{[{\text{CH}}_3\text{COOH}]}\right)$

In the solution mixture:

We have 25 mL of 0.2 M ${\text{CH}}_3\text{COONa}$ and 25 mL of 0.02 M ${\text{CH}}_3\text{COOH}$.

The concentration ratio $\left(\frac{[{\text{CH}}_3\text{COONa}]}{[{\text{CH}}_3\text{COOH}]}\right)$ becomes $\frac{25\times 0.2}{25\times 0.02}=10$.

Given that the pH of the solution is 5, we substitute into the equation:

$5={\text{pK}}_{\text{a}}+\log 10$

Since $\log 10=1$, we solve for ${\text{pK}}_{\text{a}}$:

$5={\text{pK}}_{\text{a}}+1\Rightarrow {\text{pK}}_{\text{a}}=4$

Converting from ${\text{pK}}_{\text{a}}$ to $K_{\text{a}}$, we use:

$K_{\text{a}}=10^{-{\text{pK}}_{\text{a}}}=10^{-4}$

Thus, since the dissociation constant $K_{\text{a}}$ is given as $x\times 10^{-5}$, compare:

$10^{-4}=10\times 10^{-5}$

Therefore, $x=10$.