ADV-2026-P2-CHE-Q11 · Acetic Acid Adsorption on Charcoal

01The question — verbatim

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,

x/m = k C^1/n

If the plot of log₁₀(x/m) against log₁₀C gives a straight line with slope 1, the value of k in L mol⁻¹ is ______.

Given: molar mass of acetic acid = 60 g mol⁻¹ · K_a = 1.0 × 10⁻⁵ · x = mass (g) of acetic acid adsorbed · m = mass (g) of charcoal · C = equilibrium concentration of acetic acid in solution after adsorption · k and n are constants for the acetic acid–charcoal system.

Section 3 — numerical value (+4 for the correct value, 0 otherwise, no negative marking). Run the experiment below and the answer builds itself.

02The experiment

Everything happens on this bench. Press the button, watch what changes, and read the instruments as they come alive.

A digital balance, a conical flask on a shaker, and a pH meter. Each orange dot is 0.01 g of acetic acid.

now

Prepare the solution

0.45 g of acetic acid, 50 mL of water.

Instrument panel

Nothing measured yet. Press Start the experiment.

What this experiment is doing. Charcoal is riddled with pores, so one gram carries an enormous surface. Acid molecules that bump into it stick to that surface — they are adsorbed. Nothing is destroyed: the acid simply leaves the water and sits on the solid, so the solution ends up weaker than it started.

03What the numbers say

Each line appears the moment the measurement behind it is taken — nothing is assumed ahead of the experiment.

Working

Where the 0.45 g went

The bar always adds up to the 0.45 g weighed out.

Why x/m and not just x. Two grams of charcoal would hold twice as much acid, so the raw mass adsorbed describes the amount of charcoal, not the material. Dividing by the mass of charcoal gives a number that belongs to the charcoal itself.

04The Freundlich plot

Take logarithms of the isotherm and the power law straightens into a line. Its gradient is the only thing left between the measurement and k.

Straightening the isotherm

x/m = k C^1/n

log₁₀(x/m) = log₁₀k + (1/n) log₁₀C

That is y = c + sx: the gradient s is 1/n and the intercept is log₁₀k.

the line your measurement slope triangle intercept = log k

Reading the line

gradient = 1/n

—

intercept

—

= log₁₀ k

reading the line The plot appears once x/m and C have been measured.

Why logarithms help. A power law is a curve, and eyeballing a curve tells you very little. Taking logs of both sides turns it into a straight line, and a straight line hands you two numbers at once: its gradient and where it crosses the axis.

05Result

Freundlich constant

The result appears when the experiment concludes.

the answer

—

Try it differently

The experiment is fixed by the question, but the arithmetic is not. Switch the concentration the isotherm uses and watch what happens to k.

Finish the experiment first.

Which C the isotherm means. The isotherm describes the balance the surface actually settles into, so C is the concentration left in the solution at equilibrium. The initial concentration describes a flask that no longer exists by the time the measurement is taken.

Common mistakes on this question

Lab notebook

Written as the experiment runs.

Nothing recorded yet.

Final record

06Step-by-step solution

The solution is still locked

Run the experiment above and the derivation opens by itself.
Steps completed: 0 of 7

What went into the flask

moles = 0.45 g ÷ 60 g mol⁻¹ = 0.0075 mol

C_initial = 0.0075 mol ÷ 0.050 L = 0.15 mol L⁻¹

The volume has to be in litres. Dividing by 50 instead of 0.050 gives 0.00015, which is moles per millilitre.
The pH reading is about H⁺, not about the acid

[H⁺] = 10^−pH = 10^−3.0 = 1.0 × 10⁻³ mol L⁻¹

Acetic acid is weak: only a small fraction of it has given up a proton, so the hydrogen ions the meter sees are nowhere near the amount of acid present. That conversion is the whole point of being handed K_a.
K_a turns [H⁺] into the acid concentration

CH₃COOH ⇌ CH₃COO⁻ + H⁺ K_a = [CH₃COO⁻][H⁺] / [CH₃COOH]

Both ions come only from the acid, so [CH₃COO⁻] ≈ [H⁺], and the expression collapses:

K_a = [H⁺]² / C ⟹ C = [H⁺]² / K_a

C = (1.0 × 10⁻³)² ÷ (1.0 × 10⁻⁵) = 0.10 mol L⁻¹

Sense check: 0.10 M is below the 0.15 M you started with, which is exactly what adsorption should do.
Mass balance — the acid moved, it did not vanish

n_left = C × V = 0.10 × 0.050 = 0.005 mol

mass left = 0.005 × 60 = 0.30 g

x = 0.45 − 0.30 = 0.15 g adsorbed

0.30 g in the solution plus 0.15 g on the charcoal is the 0.45 g weighed out. A third of the acid ended up on the surface.
x/m

x/m = 0.15 g ÷ 1.0 g = 0.15

x is a mass, as the question states, so x/m is grams per gram — a pure number.
The gradient gives n

log₁₀(x/m) = log₁₀k + (1/n) log₁₀C

A straight line whose gradient is 1/n. The question hands you the gradient:

1/n = 1 ⟹ n = 1 ⟹ x/m = kC
k

0.15 = k × (0.10)¹ ⟹ k = 0.15 ÷ 0.10 = 1.5

units: (g/g) ÷ (mol L⁻¹) = L mol⁻¹

The C in the isotherm is the concentration left at equilibrium, 0.10 M — not the 0.15 M you started with. Using 0.15 gives 1.0 and is the commonest wrong answer here.
Answer

k = 1.5 L mol⁻¹

Marking: +4 for 1.5, 0 otherwise. No negative marking, so there is never a reason to leave this blank.

✓ Independent verification

Derived before anything was looked up. Every step — the mole calculation, the K_a inversion, the mass balance, the gradient and the final division — was fixed from first principles, and only then compared with a published solution.
The K_a route checked in both directions: forwards it gives C = 0.10 M from pH 3.00, and substituting back, [H⁺] = √(K_aC) = 1.000 × 10⁻³, reproducing the meter reading exactly.
Conservation checked explicitly: 0.30 g in solution + 0.15 g adsorbed = 0.45 g weighed out, to machine precision. The simulation enforces the same identity frame by frame while the flask is shaking.
Two independent routes to k: dividing (x/m) by C gives 1.5, and reading the intercept of the log–log line gives log₁₀k = 0.176091, i.e. k = 1.500000. Both are shown on the plot.
Units confirmed by dimensional analysis: (g g⁻¹) ÷ (mol L⁻¹) = L mol⁻¹, which is the unit the question asks for — a check that x was correctly taken as a mass rather than a number of moles.
The traps reproduced deliberately: using the initial 0.15 M gives exactly 1.0; treating pH 3 as the acid concentration gives an absurd 150; dividing by 50 instead of 0.050 L gives 0.00015; putting x in moles gives 0.025 L g⁻¹. None is 1.5.
The approximation quantified rather than assumed: solving the weak-acid equilibrium exactly, without treating the remaining acid as wholly undissociated, gives C = 0.1010 M and k = 1.455 — a 3.0% shift. The question's round numbers are chosen so the standard approximation lands on a clean 1.5, which is the intended answer.
Answer key cross-check: a published worked solution for this paper follows the same steps and reports the same intermediate values (10⁻³ M, 0.10 M, 0.005 mol, 0.30 g, 0.15 g, n = 1) and the same answer, 1.5 L mol⁻¹. IIT Roorkee's official key was not retrievable at the time of writing.
Headless-browser run of this page: the whole run driven end to end, mass conservation asserted mid-shake, clean console, and zero horizontal overflow at 390 px and 360 px.

07Concepts and classification

Relations used

moles = mass ÷ molar mass

C = moles ÷ volume in litres

[H⁺] = 10^−pH

weak acid: K_a = [H⁺]² / C ⟹ C = [H⁺]²/K_a

mass balance: x = mass taken − mass left in solution

Freundlich: x/m = k C^1/n

log form: log(x/m) = log k + (1/n) log C

gradient = 1/n · intercept = log k

adsorption is a surface process — nothing leaves the flask

Classification

Simulation ID	ADV-2026-P2-CHE-Q11
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Physical)
Chapter	Surface Chemistry
Topic	Freundlich adsorption isotherm from solution
Subtopics	Adsorption from solution · Freundlich isotherm and its logarithmic form · Gradient and intercept of the log–log plot · Weak-acid equilibrium and K_a · pH and hydrogen-ion concentration · Mass balance on the adsorbate
Question type	Numerical value (+4 / 0, no negative marking)
Difficulty	Moderate
Answer	— run the experiment

Teacher notes

Objective. Two chapters in one question: the ionic-equilibrium step that turns a pH into a concentration, and the surface-chemistry step that turns a concentration into k. Students who know both separately still lose this one at the join.
Runs as a demo. Press Auto run and the whole experiment plays through in about half a minute — long enough to narrate, short enough to repeat.
Stop the class at the pH step. Ask for the acid concentration at pH 3 before revealing K_a. Most will say 10⁻³ M. Then let the next step find 0.10 M — a hundred times bigger.
Expected wrong answers. 1.0 by far (using the initial 0.15 M in the isotherm); then 150 (treating pH as the acid concentration); occasionally 0.025 (putting x in moles).
Discussion question. "If we doubled the charcoal to 2.0 g, what would happen to x, to x/m, and to k?" (More acid adsorbed, a lower equilibrium C, but k unchanged — it is a constant of the system.)
Board demonstration. Shaking tea with a spoonful of activated charcoal decolourises it in under a minute. Same physics, and it lands.

EXPERIMENT
COMPLETE

The acid tracked from the balance to the charcoal surface

—