Answer to Question #329812 in General Chemistry for Bwalya

Question

18. 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

standard solution, acidified with H2SO4 was put in a conical flask. A titre volume 15.50

mL of KMnO4 solution was required to reach the end-point. The titration reaction is:

Fe2+(aq) + MnO4

−(aq) → Fe3+(aq) + Mn2+(aq)

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

(ii) Calculate the molarity of the standard solution.

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

Answer 1

(i) Balance the above redox reaction in acidic medium. (Basic medium is a mistake, the solution is acidified with H₂SO₄).

Fe²⁺_(aq) + MnO⁴⁻_(aq) → Fe³⁺_(aq) + Mn²⁺_(aq)

Fe²⁺_(aq) - e^- → Fe³⁺_(aq) x5

MnO₄⁻_(aq) + 8H⁺ + 5e^- → Mn²⁺_(aq) + 4H₂O x1

5Fe²⁺_(aq) + MnO⁴⁻_(aq) + 8H⁺ → 5Fe³⁺_(aq) + Mn²⁺_(aq) + 4H₂O

(ii) Calculate the molarity of the standard solution.

c(Fe²⁺) = 19.60/392.21 = 0.05 M

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

20 mL = 0.020 L

n(Fe²⁺) = 0.05*0.020 = 0.001 mol => n(KMnO₄) = n(Fe²⁺)/5 = 0.0002 mol

15.50 mL = 0.01550 L

c(KMnO₄) = 0.0002/0.01550 =0.0129 M