Problem 28
Question
Which of the following salts, would give \(\mathrm{SO}_{2}\) with hot and dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and also decolorize \(\mathrm{Br}_{2}\) water ? (a) \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) (b) \(\mathrm{NaHSO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Na}_{2} \mathrm{~S}\)
Step-by-Step Solution
Verified Answer
The correct salt is \(\mathrm{Na}_{2} \mathrm{SO}_{3}\).
1Step 1: Understanding the Reaction with Sulfuric Acid
Sodium sulfite (\(\mathrm{Na}_{2} \mathrm{SO}_{3}\)) reacts with hot and dilute sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) to produce sulfur dioxide (\(\mathrm{SO}_{2}\)), water (\(\mathrm{H}_{2} \mathrm{O}\)), and sodium sulfate (\(\mathrm{Na}_{2} \mathrm{SO}_{4}\)). The reaction is:\[\mathrm{Na}_{2} \mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2} + \mathrm{H}_{2} \mathrm{O} + \mathrm{Na}_{2} \mathrm{SO}_{4}\]This means \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) is a potential candidate.
2Step 2: Identifying Decolorizing Agent
Sodium sulfite (\(\mathrm{Na}_{2} \mathrm{SO}_{3}\)) can also decolorize bromine water. This happens because \(\mathrm{SO}_{3}^{2-}\) ions reduce \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\), thus removing the color of bromine water, as shown:\[\mathrm{Br}_{2} + \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{Br}^{-} + \mathrm{SO}_{4}^{2-} + 2\mathrm{H}^{+}\]This confirms that \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) fulfills both conditions of the problem.
3Step 3: Analyzing Other Options
The other options (\(\mathrm{NaHSO}_{4}\), \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), \(\mathrm{Na}_{2} \mathrm{~S}\)) either do not produce \(\mathrm{SO}_{2}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) or do not decolorize bromine water. Thus, these are not suitable choices.
Key Concepts
Sulfur Dioxide ProductionDecolorization of Bromine WaterSodium Sulfite and Sulfuric Acid Reaction
Sulfur Dioxide Production
When sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) is mixed with hot and dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)), it undergoes a chemical reaction that produces sulfur dioxide (\( \text{SO}_2 \)). This reaction also results in the formation of water (\( \text{H}_2\text{O} \)) and sodium sulfate (\( \text{Na}_2\text{SO}_4 \)).
This is a classic example of how acids can react with salts to produce gases. The equation for this reaction is shown below:
\[\text{Na}_2\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_2 + \text{H}_2\text{O} + \text{Na}_2\text{SO}_4\]
Through this process, hot and dilute sulfuric acid acts as a catalyst that facilitates the release of sulfur dioxide.
This conversion is highly valued in both laboratory and industrial processes due to the vast applications of sulfur dioxide.
This is a classic example of how acids can react with salts to produce gases. The equation for this reaction is shown below:
\[\text{Na}_2\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_2 + \text{H}_2\text{O} + \text{Na}_2\text{SO}_4\]
Through this process, hot and dilute sulfuric acid acts as a catalyst that facilitates the release of sulfur dioxide.
This conversion is highly valued in both laboratory and industrial processes due to the vast applications of sulfur dioxide.
Decolorization of Bromine Water
Bromine water comes with a distinct reddish-brown color. When sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) is added to it, an interesting chemical reaction occurs. Sodium sulfite acts as a reducing agent in this scenario.
The \( \text{SO}_3^{2-} \) ions present in sodium sulfite reduce bromine (\( \text{Br}_2 \)) to bromide ions (\( \text{Br}^- \)). This change from bromine to bromide results in the disappearance of the color in the bromine water:
\[\text{Br}_2 + \text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow 2\text{Br}^- + \text{SO}_4^{2-} + 2\text{H}^+\]
This process makes sodium sulfite an effective decolorizing agent for bromine water, showcasing its ability to alter the visual properties of chemical solutions through reduction reactions.
The \( \text{SO}_3^{2-} \) ions present in sodium sulfite reduce bromine (\( \text{Br}_2 \)) to bromide ions (\( \text{Br}^- \)). This change from bromine to bromide results in the disappearance of the color in the bromine water:
\[\text{Br}_2 + \text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow 2\text{Br}^- + \text{SO}_4^{2-} + 2\text{H}^+\]
This process makes sodium sulfite an effective decolorizing agent for bromine water, showcasing its ability to alter the visual properties of chemical solutions through reduction reactions.
Sodium Sulfite and Sulfuric Acid Reaction
The interaction between sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is a well-documented reaction that illustrates the properties of sulfites and acids.
In this reaction, sulfite ions react with hydrogen ions in sulfuric acid to produce sulfur dioxide, water, and sodium sulfate.
This reaction highlights the capability of sulfites to engage in chemical reactions, releasing gases commonly utilized in industry and various chemical processes.
Understanding these interactions is crucial for students of chemistry, as it underscores key concepts about chemical reactivity and compound transformations.
In this reaction, sulfite ions react with hydrogen ions in sulfuric acid to produce sulfur dioxide, water, and sodium sulfate.
This reaction highlights the capability of sulfites to engage in chemical reactions, releasing gases commonly utilized in industry and various chemical processes.
Understanding these interactions is crucial for students of chemistry, as it underscores key concepts about chemical reactivity and compound transformations.
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