Question

You want to determine the concentration of Fe2+in a solution. You decide to determine the Fe2+ concentration by means of a redox titration with potassium permanganate. You use 50 mL of the Fe2+ solution and adjust the pH to 1. You need 26.78 mL of a 0.0179 M KMnO4solution to reach the endpoint.

a. What is the reaction that takes place? Write the balanced chemical equation.

b. What is the concentration of Fe2+ in your original solution?

c. What is the balanced chemical equation for the titration of Fe2+with potassium dichromate? What volume of potassium dichromate (c=0.0148 M) would you have needed for this titration (the pH is acidic)?

d. Provide at least two methods to determine the endpoint of the titration with permanganate. Explain exactly how you would determine the endpoint. Do not just state the method

Answer 1

a. MnO₄^- + 5Fe²⁺ + 8H⁺ ---> Mn²⁺ + 5Fe³⁺ + 4H₂O

b. You use 50 mL of the Fe2+ solution and adjust the pH to 1. You need 26.78 mL of a 0.0179 M KMnO4solution to reach the endpoint.

1 mole of MnO₄^- is needed to react with 5 mole of Fe.

No of moles of MnO₄^- needed originally = 0.0179*26.78/1000 moles

No of moles of Fe present in 50 mL solution = 5*0.0179*26.78/1000 moles = 2.39*10^-3 moles

Molarity of Fe solution = 2.39*10^-3 * 1000/50 =0.048 M

c. Cr₂O₇^2-+ 6Fe²⁺ + 14H⁺ -> 6Fe³⁺ + 2Cr³⁺ + 7H₂O

50 mL 0.048 M Fe = 2.39*10^-3 moles Fe

No of Cr₂O₇^2- needed = 2.39*10^-3/6 = 4*10^-4 moles

Volume of Cr₂O₇^2- needed = 1000 * 4*10^-4/0.0148 = 27 mL

d. 1. You can use redox titration method to determine the end-point. Here a solution of Fe²⁺ is titrated with MnO₄^- Potassium permanganate is intensely purple in solutions, whereas the products of the reaction are essentially colorless at the concentrations used, so the "endpoint" of the titration was signaled by the appearance of a pink color in the sample being titrated, as one excess drop of KMnO₄ (beyond that needed to react with the Fe²⁺ present) was added.

2. You can also determine the endpoint by potentiometric method.

E⁰_MnO4-/Mn2+ = 1.51 V

E⁰_Fe3+/Fe2+ = 0.77 V

On addition of KMnO₄, [Fe3+]/[Fe2+] will change, so potential of the system will change with progress of reaction. At the equivalence point sharp change in the potential will be observed. Draw dE/dV curve to determine the end-point. [dV = change of volume of KMnO₄ added]