Question

An acidified solution of manganate (VII) of permanganate ions, MnO4-, is a strong oxidising agent and will be used in an experiment to determine the percentage of iron in a sample of ammonium iron (II) sulfate hexahydrate. The titration is self-indicating because at the equivalence point the products are a different colour than the original reactants. The two half reactions are:

MnO4-(aq) + 8H+(aq) + 5e- ---> Mn2+(aq) + 4H2O(l)

Fe^2+(aq) ---> Fe^3+(aq) + e-

a) What is a redox titration?

b) One of the reactants in this experiment is AR ammonium iron (II) sulfate. What does AR mean?

c) What is the formula for ammonium iron (II) sulfate hexahydrate?

d) Amy dissolved 0.3101g of Na2C2O4 in 30mL of water and 15mL of 3M H2SO4. She titrated this solution with a KMnO4 solution. What was the molarity of the KMnO4 if it took 24.90mL of titrant to reach a permanent pale pink equivalence point?

Answer 1

Answer:

1) Redox titration : It is a quantitative analysis involving a redox reaction between titrant and titrand of which one is oxidant and other is reductant.

2) AR means Analytical Reagents. These reagents are used for analytical purpose and hence ensured high purity.

3) ammonium iron (II) sulfate hexahydrate

Here Iron i.e. Fe (II) i.e. +2 oxidation state. Sulfate SO₄^2- i.e. -2 charge but there is one more cationic special ammonium ion NH₄⁺ with +1 charge. If we use one NH₄⁺ ion then there will be net +1 charge on compound. But compound is neutral and hence we need to take 2 NH₄⁺ ions and 2 SO₄^2- ions which will give neutral compound with minimum possible ratio of constituent ions.

Hence, Molecular Formula is (NH₄)₂Fe(SO₄)₂·6H₂O

4) There will be a redox reaction between Na₂C₂O₄ and KMnO₄ in acidic medium. A balanced redox reaction is,

5 Na₂C₂O₄ + 2 KMnO₄ + 8 H2SO4 (aq) $\rightarrow$ K₂SO₄ (aq) + 2 MnSO₄ (s) + 8H₂O + 10CO₂ (g) + 10 Na₂SO₄ (aq).

Mole relationship for reactant is (only part of interest taken)

5 moles of Na₂C₂O₄$\equiv$ 2 moles of KMnO₄. .......... (1)

Let us calculate # of moles of  Na₂C₂O₄ .

Molar mass of Na₂C₂O₄ = 134 g.mol^-1 , Mass of  Na₂C₂O₄ titrated = 0.3101 g

# of moles of Na₂C₂O₄ = Mass titrated / Molar mass = 0.3101 g / 134 gmol^-1. = 0.0023 mol.

By eq.1 mole relationship, let us calculate # of moles of KMnO4 required till the end point.

If, 5 moles of Na₂C₂O₄$\equiv$ 2 moles of KMnO₄

then, 0.0023 moles of Na₂C₂O₄$\equiv$ say "N" moles of KMnO₄.

By cross multiplication,

5*N = 2 * 0.0023

N = 0.0046 / 5

N = 0.00092 moles.

# of moles of KMnO4 = 0.00092 mol. ............(2)

Molarity of KMnO4 = ? if Volume of KMnO4 till end point = 24.90 mL = 0.0249 L

# of moles of KMnO4 = Molarity of KMnO4 * Volume in L

So,

Molarity of KMnO4 = # of moles of KMnO4 / Volume in L

Molarity of KMnO4 = 0.00092 mol / 0.0249 L

Molarity of KMnO4 = 0.037 mol/L

Molarity of KMnO4 = 0.037 M

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