QJEEM23Ch10b4: JEE Main 2023 Chemistry - Class 11 JEEM23Ch10b4 Solution: The dissociation constant of acetic acid is . When 25 mL of 0.2 …

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 $\text{pK}_\text{a}$ :

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

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, $x = 10$ .