The molar ratio of dimer to monomer for 0.1 M acetic acid in benzene is equal to: — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let monomer concentration be [M] and dimer concentration be [D]. The equilibrium is: 2 M ⇌ D with K = [D]/[M]^2 = 1.5 × 10^2 = 150. This means [D] = 150 × [M]^2. The total analytical concentration of acetic acid is C = 0.1 M. Since each dimer contains 2 monomer units: C = [M] + 2[D] = 0.1 => [M] + 2 × (150 × [M]^2) = 0.1 => 300 [M]^2 + [M] - 0.1 = 0. Solving this quadratic equation: [M] = 1/60 M. Dimer [D] = 150 × (1/60)^2 = 1/24 M. Molar ratio of dimer to monomer = [D] / [M] = (1/24) / (1/60) = 60/24 = 5/2 = 5:2.