Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
500 mL of 0.150 M AgNO3 solution were added in 500 mL of 1.09 M Fe2+ 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 KMnO4 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]
ChemistryChemical EquilibriumJEE Main
Solution:
1758 Upvotes Verified Answer
The correct answer is: 3
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 + = 12·48 Meqs = 12·48 mM  
(Z factor= 1)

Total mM of  Fe 2 + left = 12 · 48 × 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 + = 4 6  mM 1 0 0 0  mL = 0 · 0 4 6  M

K C = Fe 3+ Ag + Fe 2+ = 0·046 0·499×0·029 =3.18 M 1

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.