To 500 ml of 0.9 M-AgNO₃ solution was added 500 ml of 1.0M-Fe²⁺ solution and the reaction is allowed — Chemical Equilibrium Chemistry Question
Question
To 500 ml of 0.9 M-AgNO₃ solution was added 500 ml of 1.0M-Fe²⁺ solution and the reaction is allowed to achieve equilibrium at 25°C. Ag⁺(aq) + Fe²⁺(aq) ⇌ Fe³⁺(aq) + Ag(s). For 30 ml of the solution, 25 ml of 0.06 M-KMnO₄ was required for oxidation of Fe²⁺ present. The equilibrium constant (in M⁻¹) for the reaction at 25°C is
💡 Solution & Explanation
Initial concentrations after mixing equal volumes: [Ag⁺] = 0.45 M, [Fe²⁺] = 0.50 M. Let x be the amount reacted at equilibrium: [Ag⁺] = 0.45 - x, [Fe²⁺] = 0.50 - x, [Fe³⁺] = x. 25 ml of 0.06 M KMnO₄ oxidizes Fe²⁺: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Moles of KMnO₄ = 25 × 0.06 = 1.5 mmol. Moles of Fe²⁺ in 30 ml = 5 × 1.5 = 7.5 mmol. Concentration of Fe²⁺ at equilibrium = 7.5 mmol / 30 ml = 0.25 M. Thus, 0.50 - x = 0.25 => x = 0.25 M. Equilibrium concentrations: [Ag⁺] = 0.20 M, [Fe²⁺] = 0.25 M, [Fe³⁺] = 0.25 M. K_c = [Fe³⁺] / ([Ag⁺][Fe²⁺]) = 0.25 / (0.20 × 0.25) = 1 / 0.20 = 5 M⁻¹. Therefore, correct answer is 5.