Answer to Question #325547 in General Chemistry for Ed123

Question

 In a redox titration, potassium permanganate, KMnO4, was standardized using a 19.60 g/L FeSO4(NH4)SO4.6H2O, (molar mass = 392.21 g/mol) standard solution. A 20.00 mL of the FeSO4(NH4)SO4.6H2O standard solution, acidified with H2SO4 was put in a conical flask. Potassium permanganate, KMnO4  was placed in the burette and the titration conducted against 20.00 mL of the FeSO4(NH4)SO4.6H2O standard solution in the conical flask.A titre volume 15.50 mL of KMnO4 solution was required to reach the end-point. The titration reaction is:

(i) Balance the above redox reaction in basic medium.

(ii) Calculate the molarity of the 19.60 g/L FeSO4(NH4)SO4.6H2O, standard solution.

(iii) Determine molarity of MnO4-(aq) in the burette.

Answer 1

(i)                 Balance the above redox reaction in basic medium.

Answer: MnO₄^- + 5Fe²⁺ + 4H₂O = Mn²⁺ + 8OH^- + 5Fe³⁺

(ii)               Calculate the molarity of the 19.60 g/L FeSO4(NH4)SO4.6H2O, standard solution.

Answer: 19.60 / 392.21 = 0.045 (mol/L)

 

(iii)              Determine molarity of MnO4-(aq) in the burette.

Answer:

Moles Fe²⁺: 0.045 * 0.02 = 9.99 x 10^-4 mol

Mole ratio is 1 : 5. So, moles MnO₄^-: 9.99 x 10^-4 / 5 = 2.0 x 10^-4 mol

2.0 x 10^-4 / 0.0155 = 0.013 M