JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Surface Chemistry

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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.

Answer: 1.5

Solution

Initial moles of CH₃COOH =

\frac{0.45}{60} = 0.0075

mol

Initial molarity =

\frac{0.0075}{50} \times 1000 = 0.15

M

After adsorption, pH = 3 ⇒ [H⁺] = 10⁻³ M

For the equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺

K_{a} = 1.0 \times 1 0^{- 5} = \frac{( 1 0 ^{- 3} ) ( 1 0 ^{- 3} )}{C - 1 0 ^{- 3}}

Solving:

C = 0.1

M (equilibrium concentration)

Mass of CH₃COOH adsorbed =

(0.15 - 0.1) \times 50 \times 60/1000 = 0.15

g

Since slope = 1, we have

n = 1

, so:

\frac{x}{m} = k C

lo g (\frac{x}{m}) = lo g k + lo g C

lo g (0.15) = lo g k + lo g (0.1)

lo g (1.5) = lo g k

Therefore,

k = 1.5

L mol⁻¹

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Surface Chemistry

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