Problem 98 - A Complete Resource Book in Chemistry for JEE Main

Question

A substance on treatment with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) liberates a colourless gas which produces (i) turbidity with baryta water and (ii) turns acidified dichromate solution green. These reactions indicate the presence of (a) \(\mathrm{CO}_{3}^{2-}\) (b) \(\mathrm{S}^{2-}\) (c) \(\mathrm{SO}_{3}^{2-}\) (d) \(\mathrm{NO}_{2}^{-}\)

Step-by-Step Solution

Answer

The substance contains \( \mathrm{SO}_{3}^{2-} \).

Step 1: Analyze the Reaction with Dilute H2SO4

When the substance is treated with dilute \( \mathrm{H}_2\mathrm{SO}_4 \), a colorless gas is liberated. This step requires identifying which ion from the options reacts with \( \mathrm{H}_2\mathrm{SO}_4 \) to produce a gas. Options given are \( \mathrm{CO}_{3}^{2-} \), \( \mathrm{S}^{2-} \), \( \mathrm{SO}_{3}^{2-} \), and \( \mathrm{NO}_{2}^{-} \). All can theoretically react with strong acids, but here focus on what colorless gas each could produce.

Step 2: Reaction with Baryta Water Test

The liberated gas creates turbidity with baryta water, \( \mathrm{Ba(OH)_2} \). This test commonly identifies \( \mathrm{CO}_2 \) because it reacts with \( \mathrm{Ba(OH)_2} \) to form \( \mathrm{BaCO}_3 \), a white precipitate. Therefore, a key clue suggests that \( \mathrm{CO}_2 \) is involved.

Step 3: Dichromate Solution Test

The dichromate solution turns green upon reaction, indicating reduction to \( \mathrm{Cr}^{3+} \). This reduction is a typical reaction when \( \mathrm{SO}_2 \) gas is present – resulting from \( \mathrm{SO}{_3}{^2-} \) reacting with \( \mathrm{H}_2\mathrm{SO}_4 \). Thus, when reduced, \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) turns to \( \mathrm{Cr}^{3+} \), producing a green solution.

Step 4: Identify the Substance

Combining the results from the baryta water and dichromate tests, recognizing that the generation of \( \mathrm{SO}_2 \) fits both test outcomes. The turbidity and green solution tests align with the reaction conditions for \( \mathrm{SO}_2 \), indicating the presence of \( \mathrm{SO}{_3}{^2-} \) which produces \( \mathrm{SO}_2 \) gas upon reaction with \( \mathrm{H}_2\mathrm{SO}_4 \).

Key Concepts

Reactions with Sulfuric Acid

Sulfuric acid, represented as \( \mathrm{H}_2\mathrm{SO}_4 \), is known for its ability to react with various substances, often liberating gases in the process. When a substance reacts with dilute sulfuric acid, it typically undergoes either a decomposition or a displacement reaction, releasing different gaseous products. In the case of carbonate ions, \( \mathrm{CO}_3^{2-} \), they react with sulfuric acid to produce carbon dioxide \( \mathrm{CO}_2 \), a colorless gas. This reaction is represented as:

• \( \mathrm{CO}_3^{2-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{SO}_4^{2-} \)

Similarly, sulfite ions, \( \mathrm{SO}_3^{2-} \), react with dilute sulfuric acid to produce sulfur dioxide \( \mathrm{SO}_2 \), which is also colorless:

• \( \mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{SO}_4^{2-} \)

In contrast, sulfide ions \( \mathrm{S}^{2-} \) will produce hydrogen sulfide \( \mathrm{H}_2\mathrm{S} \), a gas with a characteristic rotten egg smell. Understanding these reactions is essential for predicting the behavior of substances in chemical processes involving sulfuric acid.

Gas Identification Tests

Gas identification tests are crucial for determining the type of gas produced in a chemical reaction. One common test involves using baryta water \( \mathrm{Ba(OH)_2} \), which is used specifically to detect carbon dioxide. When carbon dioxide is bubbled through baryta water, it forms a white precipitate of barium carbonate:

• \( \mathrm{CO}_2 + \mathrm{Ba(OH)_2} \rightarrow \mathrm{BaCO}_3 + \mathrm{H}_2\mathrm{O} \)

This reaction results in turbidity, allowing us to confirm the presence of \( \mathrm{CO}_2 \).
Another test involves an acidified dichromate solution. This test identifies gases capable of reducing dichromate ions. Sulfur dioxide \( \mathrm{SO}_2 \) is one such gas that reduces dichromate \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) to chromium \( \mathrm{Cr}^{3+} \), turning the solution from orange to green:

• \( \mathrm{SO}_2 + \mathrm{Cr}_2\mathrm{O}_7^{2-} + 2\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + \mathrm{SO}_4^{2-} + \mathrm{H}_2\mathrm{O} \)

Such chemical tests are important tools for chemists to identify and confirm the nature of gases evolved in various reactions.

Chemical Reaction Analysis

Analyzing a chemical reaction involves understanding the substances and their interactions. In the context of reactions with sulfuric acid and the resulting gases, analyzing the reaction means predicting which gas is produced based on the initial compounds.
For instance, given the options in the exercise, understanding the properties and behaviors of carbonate \( \mathrm{CO}_3^{2-} \), sulfide \( \mathrm{S}^{2-} \), sulfite \( \mathrm{SO}_3^{2-} \), and nitrite \( \mathrm{NO}_2^{-} \) ions is key.

• Carbonates react to release \( \mathrm{CO}_2 \).
• Sulfites release \( \mathrm{SO}_2 \).
• Sulfides form \( \mathrm{H}_2\mathrm{S} \).
• Nitrites generally do not yield a colorless gas with sulfuric acid.

By systematically considering the reactions of each ion with \( \mathrm{H}_2\mathrm{SO}_4 \), we can perform deductive reasoning to identify the substance and gas involved. Utilizing typical reaction conditions and subsequent tests, like those with baryta water or dichromate solutions, aids in confirming the analysis. This methodical approach helps chemists validate their predictions about reaction outcomes.