JEE Challenger
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ChemMedAdv 2021 • Paper 2

Calculate Moles of Ferrous Ion in Solution

Comprehension Passage

An iron-containing sample weighing 5.6 g5.6\text{ g} is completely dissolved in cold, dilute HCl\text{HCl} to prepare a 250 mL250\text{ mL} solution. A 25.0 mL25.0\text{ mL} aliquot of this prepared solution requires 12.5 mL12.5\text{ mL} of 0.03 M KMnO40.03\text{ M } \text{KMnO}_4 solution to reach the titration end point. The total number of moles of Fe2+\text{Fe}^{2+} present in the 250 mL250\text{ mL} solution is given by x×10−2x \times 10^{-2} (assuming complete dissolution of FeCl2\text{FeCl}_2). The weight percentage of iron present in the sample is y%y\%.

(Assume KMnO4\text{KMnO}_4 reacts exclusively with Fe2+\text{Fe}^{2+} in the solution. Given: Molar mass of iron = 56 g mol−156\text{ g mol}^{-1})

Determine the value of xx.

Official Numerical Answer1.875

Topics & Concepts

Redox ReactionsStoichiometryMole ConceptBalancing Redox Reactions

Step-by-Step Solution

To determine the value of xx, we analyze the redox titration reaction occurring between the ferrous ion (Fe2+\text{Fe}^{2+}) and the permanganate ion (MnO4−\text{MnO}_4^-) in an acidic medium.

1. Balanced Redox Reaction and Stoichiometry

In acidic medium (dilute HCl\text{HCl}), MnO4−\text{MnO}_4^- acts as an oxidizing agent and oxidizes Fe2+\text{Fe}^{2+} to Fe3+\text{Fe}^{3+} while being reduced to Mn2+\text{Mn}^{2+}:

MnO4−+5Fe2++8H+⟶Mn2++5Fe3++4H2O\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \longrightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

From the balanced chemical equation, 1 mole1\text{ mole} of KMnO4\text{KMnO}_4 reacts with 5 moles5\text{ moles} of Fe2+\text{Fe}^{2+}.

Alternatively, using the concept of n-factors:

  • For MnO4−\text{MnO}_4^- changing to Mn2+\text{Mn}^{2+}:
    n-factor=∣+7−(+2)∣=5\text{n-factor} = |+7 - (+2)| = 5
  • For Fe2+\text{Fe}^{2+} changing to Fe3+\text{Fe}^{3+}:
    n-factor=∣+2−(+3)∣=1\text{n-factor} = |+2 - (+3)| = 1

2. Moles of KMnO4\text{KMnO}_4 Consumed in Titration

Given:

  • Molarity of KMnO4\text{KMnO}_4 solution, M=0.03 MM = 0.03\text{ M}
  • Volume of KMnO4\text{KMnO}_4 solution used, V=12.5 mL=12.5×10−3 LV = 12.5\text{ mL} = 12.5 \times 10^{-3}\text{ L}

The number of moles of KMnO4\text{KMnO}_4 used for the titration of a 25.0 mL25.0\text{ mL} aliquot is:

nKMnO4=Molarity×Volume (in L)n_{\text{KMnO}_4} = \text{Molarity} \times \text{Volume (in L)} nKMnO4=0.03 mol L−1×12.5×10−3 L=0.375×10−3 mol=3.75×10−4 moln_{\text{KMnO}_4} = 0.03 \text{ mol L}^{-1} \times 12.5 \times 10^{-3} \text{ L} = 0.375 \times 10^{-3} \text{ mol} = 3.75 \times 10^{-4} \text{ mol}

3. Moles of Fe2+\text{Fe}^{2+} in the 25.0 mL25.0\text{ mL} Aliquot

Using the stoichiometric ratio (1 mol KMnO4≡5 mol Fe2+1\text{ mol } \text{KMnO}_4 \equiv 5\text{ mol } \text{Fe}^{2+}):

nFe2+(25 mL)=5×nKMnO4n_{\text{Fe}^{2+}}^{(25\text{ mL})} = 5 \times n_{\text{KMnO}_4} nFe2+(25 mL)=5×(3.75×10−4 mol)=1.875×10−3 moln_{\text{Fe}^{2+}}^{(25\text{ mL})} = 5 \times (3.75 \times 10^{-4} \text{ mol}) = 1.875 \times 10^{-3} \text{ mol}

4. Total Moles of Fe2+\text{Fe}^{2+} in the 250 mL250\text{ mL} Solution

Since the original solution has a total volume of 250 mL250\text{ mL}, the total number of moles of Fe2+\text{Fe}^{2+} in the entire solution is:

nFe2+(total)=nFe2+(25 mL)×250 mL25.0 mLn_{\text{Fe}^{2+}}^{(\text{total})} = n_{\text{Fe}^{2+}}^{(25\text{ mL})} \times \frac{250\text{ mL}}{25.0\text{ mL}} nFe2+(total)=1.875×10−3 mol×10=1.875×10−2 moln_{\text{Fe}^{2+}}^{(\text{total})} = 1.875 \times 10^{-3} \text{ mol} \times 10 = 1.875 \times 10^{-2} \text{ mol}

5. Finding the Value of xx

Given that the total number of moles of Fe2+\text{Fe}^{2+} in the 250 mL250\text{ mL} solution is represented by x×10−2x \times 10^{-2}:

x×10−2=1.875×10−2x \times 10^{-2} = 1.875 \times 10^{-2}

x=1.875x = 1.875

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