Question

Balanced redox equation: MnO₄^1- + 8H¹⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

	Titration 1	Titration 2
Volume of FeSO4	10 ml	10 ml
Molarity of FeSO4	0.1	0.1
Coefficient of FeSO4, from redox equation	5	5
Initial level of MnO41-	0	15.1
Final level of MnO41-	15.1	30.3
Volume of MnO41- used	15.1	15.2
Coefficient of MnO41- from redox equation	1	1
Calculated Molarity of MnO41-

Calculate the average concentration of KMnO₄ titrant using M₁xV₁xE₁ = M₂ xV₂ xE₂

Average Molarity of MnO₄^1-                                           __________

Answer 1

From , the balanced equation

mmole ratio of  Fe²⁺  and MnO₄^- = 5 : 1

  mmoles of MnO₄^- = (mmoles of Fe²⁺)/5

now, mmoles of Fe²⁺ = molarity volume

	Trial 1	Trial 2
mmoles of Fe2+	0.110 = 1	0.110 =1
mmoles of MnO4-	(1/5) = 0.2	(1/5) = 0.2
volume of MnO4- used	15.1	15.2
molarity of MnO4- = ( mmoles/volume)	(0.2/15.1) = 0.01325 M	(0.2/15.2) = 0.01316 M

average molarity KMnO₄

= ( 0.01325 + 0.01316)/2

= 0.01321 M.

or, it can be calculated by using the equation

M₁V₁E₁ = M₂V₂E₂

M₁ is molarity of Fe²⁺

M₂ is molarity of MnO₄^-  

V₁ is volume of Fe²⁺

V₂ is volume of MnO₄^-  

E₁ is equivalence factor of Fe²⁺ = 1

E₂ is equivalence factor of MnO₄^- = 5

Equivalence factor means number of electrons lost or gained.

	Trial 1	Trial 2
molarity of MnO4- (M2) = (M1V1E1/V2E2	(0.1101)/(15.15) = 0.01325 M	(0.1101)/(15.25) = 0.01316 M

Then,

average molarity KMnO₄

= ( 0.01325 + 0.01316)/2

= 0.01321 M.