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← JEE Advanced 2026 · Paper 2
ChemistryNumerical+4Question 47 of 54

JEE Advanced 2026 · Paper 2, Chemistry question 11

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,
xm=kC1/n\frac xm=kC^{1/n}
If the plot of log⁡10(x/m)\log_{10}(x/m) against log⁡10C\log_{10}C gives a straight line with slope 1, the value of kk in L mol−1^{-1} is ____.

Given: The molar mass of acetic acid is 60 g mol−1^{-1}.
The acid dissociation constant of acetic acid is 1.0×10−51.0\times10^{-5} at the given temperature.
xx is the mass (in grams) of acetic acid adsorbed.
mm is the mass (in grams) of charcoal.
CC is the equilibrium concentration of acetic acid in the solution after the adsorption is complete.
kk and nn are constants for acetic acid-charcoal system at the given temperature.

Answer and worked solution

Official answer: 1.45

Equilibrium concentration. At pH 3, [HX+]=10−3=KaC[\ce{H+}]=10^{-3}=\sqrt{K_aC}, so
C=10−610−5=0.10 M.C=\frac{10^{-6}}{10^{-5}}=0.10\text{ M}.
(Solving exactly, C=0.10+0.001=0.101C=0.10+0.001=0.101 M.)

Amount adsorbed. Initially 0.4560=7.5×10−3\frac{0.45}{60}=7.5\times10^{-3} mol in 50 mL, i.e. 0.15 M. The loss is 0.05 M×0.050 L=2.5×10−30.05\text{ M}\times0.050\text{ L}=2.5\times10^{-3} mol, so x=0.15x=0.15 g and xm=0.15\frac xm=0.15.

Freundlich. Slope 1n=1\frac1n=1, so k=x/mC=0.150.10=1.5k=\dfrac{x/m}{C}=\dfrac{0.15}{0.10}=1.5 L mol⁻¹. (With C=0.101C=0.101 M, k≈1.46k\approx1.46.)