500 mL of 0.150 M AgNO 3 solution were added in 500 mL of 1.09 M Fe 2 + solution and the reaction is allowed to reach an equilibrium at 2 5 ∘ C . Ag + aq + Fe 2 + aq ⇌ Fe 3 + aq + Ag s For 25 mL of the solution, 30 mL of 0.0832 M KMnO 4 were required for oxidation. Calculate equilibrium constant (M -1 ) of the reaction at 2 5 ∘ C . [Report your answer by rounding it upto nearset whole number]

Step-by-step solution

Ag + aq + Fe 2 + aq ⇌ Fe 3 + aq + Ag s Initial 7 5 mM 5 4 5 mM 0 0 Reaction - 4 6 mM - 4 6 mM + 4 6 mM + 4 6 mM --------------- --------------- --------------- --------------- Final 2 9 mM 4 9 9 mM 4 6 mM 4 6 mM m.eqs of KMnO 4 = 3 0 × 0 · 0 8 3 2 × 5 = 1 2 · 4 8 m.eqs (Z factor =5) m.eqs of Fe 2 + = 1 2 · 4 8 Meqs = 1 2 · 4 8 mM (Z factor= 1) Total mM of Fe 2 + left = 1 2 · 4 8 × 4 0 = 4 9 9 mM At Equilibrium concentrations Ag + = 2 9 mM 1 0 0 0 mL = 0 · 0 2 9 M Fe 2 + = 4 9 9 mM 1 0 0 0 mL = 0 · 4 9 9 M Fe 3 + =