JEE Advanced 2026 — Paper 2 — CHEM-III solution

JEE Advanced2026Paper 2CHEM-III

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 $lo g_{10} (x / m)$ against $lo g_{10} C$ 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 $1.0 \times 1 0^{- 5}$ 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.

Your answer

Solution

Initial molarity of acetic acid. Moles $= 0.45/60 = 0.0075$ in 50 mL gives $0.15$ M.

Equilibrium concentration from pH. pH $= 3 \Rightarrow [H^{+}] = 1 0^{- 3}$ M. For dissociation CH $_{3}$ COOH $⇌$ CH $_{3}$ COO $^{-}$ + H $^{+}$ ,

$K_{a} = \frac{[ H ^{+} ] ^{2}}{C - [ H ^{+} ]} = 1.0 \times 1 0^{- 5} \Rightarrow \frac{1 0 ^{- 6}}{C - 1 0 ^{- 3}} = 1 0^{- 5} \Rightarrow C - 1 0^{- 3} = 0.1 \Rightarrow C \approx 0.1$ M.

Mass adsorbed. Concentration drop is $(0.15 - 0.1) = 0.05$ M over $50$ mL gives $0.05 \times 0.05 = 0.0025$ mol $= 0.15$ g. So $x / m = 0.15/1.0 = 0.15$ .

Apply Freundlich (slope $= 1/ n = 1$ ). $lo g (x / m) = lo g k + lo g C$ :

$lo g 0.15 = lo g k + lo g 0.1 \Rightarrow lo g k = lo g 0.15 - lo g 0.1 = lo g 1.5 \Rightarrow k = 1.5$ L mol $^{- 1}$ .

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