JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

Given the pH of the resulting solution is 3.0, the concentration of H^+ ions is: [H^+] = 10^ -3 M For the dissociation of acetic acid (CH_3COOH CH_3COO^- + H^+), the acid dissociation constant is given by: K_a = [CH_3COO^-][H^+] [CH_3COOH] Assuming [CH_3COO^-] [H^+] and the equilibrium concentration of acetic acid is C, we can use the approximation: K_a = [H^+]^2 C 1.0 10^ -5 = (10^ -3 )^2 C C = 10^ -6 1.0 10^ -5 = 0.1 M The number of moles of acetic acid remaining in the 50 mL (0.05 L ) solution at equilibrium is: n_ eq = C V = 0.1 mol L ^ -1 0.05 L = 0.005 mol The mass of acetic acid remaining in the solution is: W_ eq = n_ eq Molar mass = 0.005 mol 60 g mol ^ -1 = 0.30 g The initial mass of acetic acid was 0.45 g . The mass of acetic acid adsorbed (x) by the charcoal is: x = 0.45 g - 0.30 g = 0.15 g The Freundlich adsorption isotherm is given by: x m = k C^ 1/n Taking the logarithm on both sides: _ 10 ( x m ) = _ 10 k + 1 n _ 10 C The plot of _ 10 (x/m) against _ 10 C is a straight line with a slope of 1. Therefore: 1 n = 1 n = 1 Substitute the values into the isotherm equation (m = 1.0 g ): 0.15 1.0 = k (0.1)^1 k = 0.15 0.1 = 1.5 L mol ^ -1 Answer: 1.5