Problem 46 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An aqueous solution containing \(0.10 \mathrm{~g} \mathrm{KIO}_{3}\) (formula weight \(=214.0\) ) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated \(\mathrm{I}_{2}\) consumed \(45.0 \mathrm{~mL}\) of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution. [1998-5 Marks]

Step-by-Step Solution

Answer

The molarity of the sodium thiosulphate solution is 0.0623 M.

Step 1: Write the Balanced Chemical Equations

The reaction of \(\mathrm{KIO}_{3} + 5 \mathrm{KI} + 6\mathrm{HCl} \rightarrow 3\mathrm{I}_2 + 6\mathrm{KCl} + 3\mathrm{H}_2\mathrm{O}\)describes how iodine is liberated. The iodine then reacts with sodium thiosulphate as follows: \(\mathrm{I}_2 + 2 \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2 \mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6\).

Step 2: Calculate Moles of \( \mathrm{KIO}_3 \)

First, calculate the moles of \(\mathrm{KIO}_3\) in the solution using its molecular weight:\[\text{Moles of } \mathrm{KIO}_3 = \frac{0.10 \text{ g}}{214.0 \text{ g/mol}} = 4.673 \times 10^{-4} \text{ mol}\].

Step 3: Determine Moles of Liberated \( \mathrm{I}_2 \)

From the balanced chemical equation, 1 mole of \(\mathrm{KIO}_3\) produces 3 moles of \(\mathrm{I}_2\). Therefore, calculate the moles of \(\mathrm{I}_2\) liberated:\[\text{Moles of } \mathrm{I}_2 = 3 \times 4.673 \times 10^{-4} \text{ mol} = 1.402 \times 10^{-3} \text{ mol}\].

Step 4: Calculate Moles of \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \) Used

According to the reaction, 1 mole of \(\mathrm{I}_2\) reacts with 2 moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\). Thus, calculate:\[\text{Moles of } \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 = 2 \times 1.402 \times 10^{-3} \text{ mol} = 2.804 \times 10^{-3} \text{ mol}\].

Step 5: Calculate Molarity of \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \) Solution

Molarity is defined as moles per liter. The volume of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) used was 45.0 mL, or 0.045 L. Calculate the molarity:\[M = \frac{2.804 \times 10^{-3} \text{ mol}}{0.045 \text{ L}} = 0.0623 \text{ M}\].

Key Concepts

Iodine Liberation

Iodine liberation in this context refers to a chemical reaction that involves converting iodate ions to iodine molecules. When handling chemical reactions involving iodine, you commonly see potassium iodate (\(\mathrm{KIO}_3\)) being mixed with potassium iodide (\(\mathrm{KI}\)) in an acidic solution, such as hydrochloric acid (HCl). These reactants produce iodine (\(\mathrm{I_2}\)) along with other products according to the balanced chemical equation:
- \(\mathrm{KIO}_3 + 5\mathrm{KI} + 6\mathrm{HCl} \rightarrow 3\mathrm{I}_2 + 6\mathrm{KCl} + 3\mathrm{H}_2\mathrm{O}\)
The liberation process involves using an oxidizing agent, i.e., \(\mathrm{KIO}_3\), to produce iodine from iodide ions. The liberated iodine is crucial as it can further react with other compounds in subsequent reactions, such as thiosulphate, forming a basis for iodine titration experiments. This type of reaction helps in measuring various concentrations in solution, especially in analytical chemistry.
Understanding iodine liberation forms the foundation for accurate quantitative analysis in titration processes.

Thiosulphate Titration

Thiosulphate titration is an analytical technique used to determine the concentration of certain substances through the reaction of sodium thiosulphate (\(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\)) with iodine (\(\mathrm{I}_2\)). The titration relies on iodine's ability to form a color complex with starch, turning the solution blue. Sodium thiosulphate reacts with this iodine in a reliable and consistent manner, which allows quantification of the iodine.
In this particular titration process, the chemical equation that showcases the reaction is:
- \(\mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2\mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6\)
As you add the thiosulphate solution, the reddish-brown iodine color begins to disappear. The endpoint of titration is reached once the solution turns clear, indicating all iodine has reacted.Using the balanced equation allows you to determine how much thiosulphate is needed to react with a known amount of iodine. Then, knowing the initial amount of iodine, you can calculate the molarity of the sodium thiosulphate solution. This process is essential in applications like determining the concentration of oxidizing agents.

Balanced Chemical Equations

Balanced chemical equations are vital to understanding chemical reactions, highlighting the relationship between reactants and products. Each side of a chemical equation must have the same number of each type of atom, respecting the law of conservation of mass.In our example problem, two balanced equations guide the analysis: the liberation of iodine and thiosulphate reaction. For iodine liberation:
- \(\mathrm{KIO}_3 + 5\mathrm{KI} + 6\mathrm{HCl} \rightarrow 3\mathrm{I}_2 + 6\mathrm{KCl} + 3\mathrm{H}_2\mathrm{O}\)
Here, iodate ions yield iodine through oxidation by reacting with iodide ions and acid. The balanced nature of the equation ensures that there is conservation of mass and balance in electron transfer.For the thiosulphate titration:
- \(\mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2\mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6\)
Here, iodine reacts with thiosulphate in a straightforward stoichiometry, involving electron transfer leading to distinct products.Balancing such reactions involves ensuring the side of the equation with reactants and products involves equal atoms of elements, reflecting real physical interactions that occur in the laboratory.