Question

If a titration is performed using 42.6 mL of iron (II) sulfite and 35.6 mL of 0.00265 M potassium permanganate solution, what is the molarity of the iron (II) sulfite solution used? How many grams of iron (II) sulfite are titrated in this fashion?

Answer 1

The equation for redox titration of Fe(II) with KMnO4 is :

5Fe²⁺ + MnO₄^- + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

In the equation, 5 moles of Fe(II) is titrated against 1M of MnO₄^-

​Using the formula: M₁V₁/n₁ = M₂V₂/n₂

^​​Concentration of Fe(II) being titrated against 0.00265M of MnO₄^- =( 5* 0.00265M * 35.6 )/ 42.6 = 0.0111 M

Since 1 mole of FeSO3 = 136 g, hence 0.01325 moles will contain (0.0111 * 136 =) 1.509 g of Iron (II) Sulphite in 1 litre.

Since the titrated volume is 35.6 ml, hence amount of Fe(II) will be 0.054g.