Problem 47 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A \(3.00 \mathrm{~g}\) sample containing \(\mathrm{Fe}_{3} \mathrm{O}_{4}, \mathrm{Fe}_{2} \mathrm{O}_{3}\) and an inert impure substance, is treated with excess of KI solution in presence of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The entire iron is converted into \(\mathrm{Fe}^{2+}\) along with the liberation of iodine. The resulting solution is diluted to \(100 \mathrm{~mL}\). A \(20 \mathrm{~mL}\) of the diluted solution requires \(11.0 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution to reduce the iodine present. A \(50 \mathrm{~mL}\) of the diluted solution, after complete extraction of the iodine requires \(12.80 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{KMnO}_{4}\) solution in dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) medium for the oxidation of \(\mathrm{Fe}^{2+}\). Calculate the percentages of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) in the original sample. |1996 - 5 Marksl

Step-by-Step Solution

Answer

The original sample contains 85.33% \( \mathrm{Fe}_2\mathrm{O}_3 \) and 14.67% \( \mathrm{Fe}_3\mathrm{O}_4 \).

Step 1: Calculate Iodine Moles in First Titration

First, determine the moles of iodine from the first titration. For the first titration, apply the formula for moles: \( M \times V = ext{moles} \). The moles of \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \) used is \( 0.5 \times \frac{11.0}{1000} = 0.0055 \text{ moles} \). The reaction between \( \mathrm{S}_2\mathrm{O}_3^{2-} \) and iodine is \( 2 \mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-} + 2 \mathrm{I}^- \). Hence, moles of \( \mathrm{I}_2 \) is \( 0.0055 / 2 = 0.00275 \text{ moles} \) in 20 mL.

Step 2: Calculate Total Moles of Iodine

Since the titration was done on a 20 mL sample, scale up to the whole 100 mL solution. The total moles of \( \mathrm{I}_2 \) in the complete 100 mL solution would be: \( 0.00275 \times \frac{100}{20} = 0.01375 \text{ moles}\).

Step 3: Calculate Moles of Fe in the Original Solution

The iodine is solely from the reaction of iron oxides with \( \mathrm{KI} \). The reaction for \( \mathrm{Fe}_3\mathrm{O}_4 \) and \( \mathrm{Fe}_2\mathrm{O}_3 \) with \( \mathrm{I}_2 \) involve converting all \( \mathrm{Fe}^{3+} \) to \( \mathrm{Fe}^{2+} \). From the titration, all \( \mathrm{I}_2 \) moles indicate \( \mathrm{Fe}^{3+} \) that were in the state to begin with, so moles of \( \mathrm{Fe}^{2+} \) are also \( 0.01375 \text{ moles}.\)

Step 4: Calculate moles of KMnO₄ in Second Titration

In the second titration, find the moles of \( \mathrm{KMnO}_4 \) that reacts with \( \mathrm{Fe}^{2+} \). The moles of \( \mathrm{KMnO}_4 \) are \( 0.25 \times \frac{12.8}{1000} = 0.0032 \text{ moles}.\) This represents moles of \( \mathrm{Fe}^{2+} \) in 50 mL of solution. The reducing reaction is \( 5 \mathrm{Fe}^{2+} + \mathrm{MnO}_4^- + 8 \mathrm{H}^+ \rightarrow 5 \mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4 \mathrm{H}_2\mathrm{O} \), meaning 5 moles of \( \mathrm{Fe}^{2+} \) per mole of \( \mathrm{KMnO}_4 \). So the moles of \( \mathrm{Fe}^{2+} \) for 50 mL are \( 0.0032 \times 5 = 0.016 \text{ moles}.\)

Step 5: Calculate Adjustment for 100 mL

Since the second titration was done on a 50 mL portion, scale up those \( \mathrm{Fe}^{2+} \) moles to the whole 100 mL. moles of \( \mathrm{Fe}^{2+} \) = \(0.016 \times \frac{100}{50} = 0.032\).

Step 6: Determine Excess Fe

From both titrations, you obtained the moles of \( \mathrm{Fe}^{2+} \) from \( \mathrm{I}_2 \) (0.01375) and from \( \mathrm{KMnO}_4 \) (0.032). Therefore, the moles of \( \mathrm{Fe}^{3+} \) originally present to be reduced and measured were 0.032 moles.

Step 7: Calculate Mass and Percentage of Fe₂O₃

Knowing that each mole of \( \mathrm{Fe}_2\mathrm{O}_3 \) has 2 moles of \( \mathrm{Fe} \), find the moles of \( \mathrm{Fe}_2\mathrm{O}_3 \) which is \( 0.032 / 2 = 0.016 \text{ moles}. \) Its molar mass is \( 159.7 \text{ g/mol}\), so mass is \(0.016 \times 159.7 = 2.56 \text{ grams}\). The percentage of \( \mathrm{Fe}_2\mathrm{O}_3 \) is \(\frac{2.56}{3.00} \times 100\% = 85.33\% \).

Step 8: Calculate Mass and Percentage of Fe₃O₄

Subtract above to obtain mass of \( \mathrm{Fe}_3\mathrm{O}_4 \); you already measured the sum, and the mass is 3 g, so 3 - 2.56 = 0.44 g \( \mathrm{Fe}_3\mathrm{O}_4 \). Knowing its formula weight is 231.533 g/mol, percentage \( \mathrm{Fe}_3\mathrm{O}_4 \) means \( (0.44 / 3) \times 100 = 14.67\% \).

Step 9: Conclusion

Summarize that the sample is 85.33% \( \mathrm{Fe}_2\mathrm{O}_3 \) and 14.67% \( \mathrm{Fe}_3\mathrm{O}_4 \).

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between two substances. This concept is central to iodometric titration, the method used in the given exercise to determine iron oxide content. The redox reactions involve changes in the oxidation state of chemical species.

In the context of iodometric titration:

• Iron in the sample, initially in \( ext{Fe}_2 ext{O}_3\) and \( ext{Fe}_3 ext{O}_4\), is converted into \( ext{Fe}^{2+}\) ions using KI and \( ext{H}_2 ext{SO}_4\). Iodine (\( ext{I}_2\)) is liberated in the process, indicating the success of the reaction.
• The iodine formed is then titrated with sodium thiosulfate (\( ext{Na}_2 ext{S}_2 ext{O}_3\)). In this redox reaction, thiosulfate reduces iodine to iodide ions, while itself gets oxidized to tetrathionate.
• Further titration with permanganate (\( ext{KMnO}_4\)) allows for measurement of total iron by oxidizing \( ext{Fe}^{2+}\) back to \( ext{Fe}^{3+}\).

The essence of redox reactions in iodometric titration lies in their predictable stoichiometry, which allows for precise calculations of substance concentrations based on electron transfers.

Iron Oxides

Iron oxides are chemical compounds composed of iron and oxygen. In this exercise, we deal specifically with \( ext{Fe}_2 ext{O}_3\) (iron(III) oxide) and \( ext{Fe}_3 ext{O}_4\) (magnetite). Understanding their properties and reactions is crucial to determine their proportion in a sample.

Key characteristics of these iron oxides are:

• Fe₂O₃: Also known as hematite, it is a red-brown compound where iron is in the +3 oxidation state. In redox reactions, it is reduced to \( ext{Fe}^{2+}\).
• Fe₃O₄: A black mineral composed of both iron(II) and iron(III) ions. In the exercise, iron from both states is converted to \( ext{Fe}^{2+}\) using a reducing agent.

The differing oxygen and iron stoichiometry affects how these oxides react and are measured during titration. The experiment seeks to determine how much of each oxide is present based on their ability to be reduced, representing subtle differences in their chemical reactivity and properties.

Analytical Chemistry

Analytical chemistry is the study of the composition of matter, using various techniques to quantify and analyze components in a sample. This exercise utilizes iodometric titration, a renowned method in analytical chemistry, to evaluate iron oxide content in a sample.

Key analytical steps include:

• Sample treatment: The iron oxide sample is reacted with KI in the presence of sulfuric acid to ensure a complete redox reaction that converts all iron to a detectable form. Iodine liberated in this step is key for subsequent titrations.
• Titration: By titrating iodine with \( ext{Na}_2 ext{S}_2 ext{O}_3\), the chemist quantifies iodine thoroughly, indicating how much iron was reduced initially from iron oxides. This closely many correlates with the sample's total oxidizable iron content.
• Data interpretation: Permanganate titration provides a quantitative measure of remaining iron as \( ext{Fe}^{2+}\), helping to differentiate between the two types of iron oxides.

Through such calculated titrations and understanding chemically-induced reactions, analytical chemistry aids in deciphering composition, thus addressing real-world chemical analysis challenges.