JEE Advanced 2026 Paper 2 Chemistry Question 11 Solution

At a given temperature, $0.45$ g of acetic acid in $50$ mL of water is shaken with $1.0$ g of charcoal and the pH of the resulting solution is $3.0$ . Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm,

$\frac{x}{m} = k C^{1/n}$

If the plot of $\log_{10}(x/m)$ against $\log_{10}C$ gives a straight line with slope $1$ , the value of $k$ in $L mol^{-1}$ is ____.

Given: The molar mass of acetic acid is $60$ g $mol^{-1}$ .

The acid dissociation constant of acetic acid is $1.0 \times 10^{-5}$ at the given temperature.

$x$ is the mass (in grams) of acetic acid adsorbed.

$m$ is the mass (in grams) of charcoal.

$C$ is the equilibrium concentration of acetic acid in the solution after the adsorption is complete.

$k$ and $n$ are constants for acetic acid-charcoal system at the given temperature.

From the Freundlich isotherm $\log(x/m) = \log k + (1/n) \log C$ , the slope is $1/n$ . Given slope $= 1$ , so $n=1$ .
Calculate equilibrium concentration $C$ :

The final pH is $3.0$ , so $[H^+] = 10^{-3}$ M.

For a weak acid, $[H^+] = \sqrt{K_a \cdot C}$ . $10^{-3} = \sqrt{10^{-5} \cdot C} \implies 10^{-6} = 10^{-5} \cdot C \implies C = 0.1 \text{ mol L}^{-1}$

Calculate mass of acetic acid adsorbed ( $x$ ): Initial moles $= \frac{0.45}{60} = 0.0075$ mol.

Final moles in solution ( $50$ mL) $= C \times V = 0.1 \times 0.050 = 0.005$ mol.

Moles adsorbed $= 0.0075 - 0.005 = 0.0025$ mol.

Mass adsorbed $x = 0.0025 \times 60 = 0.15$ g.

Calculate $k$ : Using $\frac{x}{m} = k C^1$ with $x = 0.15$ g, $m = 1.0$ g, and $C = 0.1$ mol/L:

$\frac{0.15}{1.0} = k (0.1) \implies k = \frac{0.15}{0.1} = 1.5 \text{ L mol}^{-1}$

Question 11

Detailed Solution

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