To 250 mL of 1.0 M AgNO 3 solution was added 250 mL of 1.0 M Fe 2+ solution and the reaction is allowed to achieve equilibrium at 25°C. Ag + (aq) + Fe 2+ (aq) ⇌ Fe 3+ (aq) + Ag(s) For 40 mL of the sol

Question

To 250 mL of 1.0 M AgNO 3 solution was added 250 mL of 1.0 M Fe 2+ solution and the reaction is allowed to achieve equilibrium at 25°C. Ag + (aq) + Fe 2+ (aq) ⇌ Fe 3+ (aq) + Ag(s) For 40 mL of the solution, 20 mL of 0.1 M KMnO 4 was required for oxidation of Fe 2+ present. The equilibrium constant (in M -1 ) for the reaction at 25°C is

TestHub · Accepted Answer

After mixing (500 mL): [Ag + ] 0 = 0.50 M and [Fe 2+ ] 0 = 0.50 M MnO 4 - used = (20)(0.1) = 2.0 mmol MnO 4 - + 5Fe 2+ + 8H + → Mn 2+ + 5Fe 3+ + 4H 2 O, so Fe 2+ = 10 mmol in 40 mL [Fe 2+ ] eq = 10/40 = 0.25 M Fe 2+ reacted = 0.50 - 0.25 = 0.25 M, so [Fe 3+ ] = 0.25 M and [Ag + ] = 0.50 - 0.25 = 0.25 M K c = [Fe 3+ ]/([Ag + ][Fe 2+ ]) = 0.25/[(0.25)(0.25)] K c = 4 M -1

Chemistry - Chemical Equilibrium Question with Solution | TestHub

Doubts & Discussion