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Chapter 21: Problem 1

Reaction of an inorganic sulphite \(X\) with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) generates compound Y. Reaction of \(\mathrm{Y}\) with NaOH gives \(\mathrm{X}\). Further, the reaction of \(X\) with \(Y\) and water affords compound \(Z . Y\) and \(Z\), respectively, are: [Main Sep. 06, 2020 (II)] (a) \(\mathrm{SO}_{2}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) (b) \(\mathrm{SO}_{3}\) and \(\mathrm{NaHSO}_{3}\) (c) \(\mathrm{SO}_{2}\) and \(\mathrm{NaHSO}_{3}\) (d) \(\mathrm{S}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{2}\)

Short Answer

Expert verified
Option (c): \(\mathrm{SO}_2\) and \(\mathrm{NaHSO}_3\).

Step by step solution

01

Analyze Reaction of X with H2SO4

The question states that an inorganic sulfite \(X\) reacts with dilute \(\mathrm{H}_2\mathrm{SO}_4\) to produce compound \(Y\). Considering the chemical properties of sulfites, \(X\) is likely \(\mathrm{Na}_2\mathrm{SO}_3\). The reaction is as follows: \(\mathrm{Na}_2\mathrm{SO}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4\). \(Y\) is \(\mathrm{SO}_2\).
02

Analyzing Reaction of Y with NaOH

Now, \(Y\) (which is \(\mathrm{SO}_2\)) reacts with \(\mathrm{NaOH}\) to yield \(X\) again. The reaction is: \(\mathrm{SO}_2 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O}\). This confirms that \(X\) is indeed \(\mathrm{Na}_2\mathrm{SO}_3\).
03

Analyzing Reaction of X with Y and Water to Form Z

\(X = \mathrm{Na}_2\mathrm{SO}_3\) and \(Y = \mathrm{SO}_2\). When these react in the presence of water, the reaction provides compound \(Z\). This reaction is: \(\mathrm{Na}_2\mathrm{SO}_3 + \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaHSO}_3\), which indicates that \(Z\) is \(\mathrm{NaHSO}_3\).
04

Identifying Compounds Y and Z

From the reactions, it's clear \(Y = \mathrm{SO}_2\) and \(Z = \mathrm{NaHSO}_3\). Therefore, based on the options given, we have: \(Y\) is \(\mathrm{SO}_2\), and \(Z\) is \(\mathrm{NaHSO}_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulfite Reaction
Inorganic sulfites, like sodium sulfite (Na鈧係O鈧�), often engage in chemical reactions with acids. Such reactions are prominent in chemistry due to their ability to release gases. Sulfite ions (SO鈧兟测伝) are known for their reductive properties.
This is why, when sodium sulfite reacts with dilute sulfuric acid (H鈧係O鈧�), sulfur dioxide (SO鈧�) gas and water are produced. The equation governing this process is:
  • Na鈧係O鈧� + H鈧係O鈧� 鈫� 2SO鈧� + 2H鈧侽 + Na鈧係O鈧�
This transformation illustrates the well-known sulfite reaction where a sulfite compound decomposes to release sulfur dioxide gas upon exposure to a strong acid.
The release of SO鈧� is a characteristic feature in reactions involving sulfites, often resulting in the restoration of the original sulfite via subsequent reactions.
Sulfur Dioxide
Sulfur dioxide (SO鈧�) is a simple, yet pivotal molecule in inorganic chemistry. This gas is colorless, with a distinct, penetrating odor recognizable by many as the smell of burnt matches.
It plays a major role in chemical reactions, particularly when it acts as a reducing agent. In the context of our exercise, the sulfur dioxide generated from sulfite reactions can further engage in reactions of its own, such as with sodium hydroxide (NaOH).
When sulfur dioxide is bubbled through sodium hydroxide solution, sodium sulfite can be regenerated:
  • SO鈧� + 2NaOH 鈫� Na鈧係O鈧� + H鈧侽
This reaction is crucial because it illustrates how SO鈧� can revert to its original sulfite form. Consequently, this cycle of reactions highlights the reversible nature of these chemical processes in inorganic chemistry.
Sodium Bisulfite
Sodium bisulfite, commonly referred to as NaHSO鈧�, is a compound with significant applications in both industrial and laboratory settings. It acts as a preservative, antioxidant, and reducing agent in various scenarios.
In our exercise, sodium bisulfite emerges from the interaction of sodium sulfite, sulfur dioxide, and water. The chemical equation representing this reaction is:
  • Na鈧係O鈧� + SO鈧� + H鈧侽 鈫� 2NaHSO鈧�
This reaction exemplifies how sodium bisulfite forms in environments rich with sulfur dioxide and moisture.
Sodium bisulfite's ability to form through the reaction of sulfite and SO鈧� in water illustrates the compound's adaptive nature, influenced by the surrounding chemical environment. This feature makes sodium bisulfite a valuable component in many biochemical and industrial processes.

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Most popular questions from this chapter

The acid having \(\mathrm{O}-\mathrm{O}\) bond is [2004S] (a) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\) (c) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\) Page \(543 / 1,773-\mathbb{Q}\) ? (d) \(\mathrm{H}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\)

Give reasons for the following : (i) The experimentally determined \(\mathrm{N}-\mathrm{F}\) bond length in \(\mathrm{NF}_{3}\) is greater than the sum of the single covalent bond radii of \(\mathrm{N}\) and \(\mathrm{F}\). [1995 - 2 Marks] (ii) Ammonium chloride is acidic in liquid ammonia solvent. [1991 - 1 Mark] (iii) Phosphine has lower boiling point than ammonia. [1989-1 Mark] (iv) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a dibasic acid. [1989-1 Mark] (v) Orthophosphorus acid is not tribasic acid. [1987-1 Mark] (vi) A bottle of liquor ammonia should be cooled before opening the stopper. [1983-1 Mark] (vii) Orthophosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), is tribasic, but phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), is dibasic. [1982 - 1 Mark]

Chlorine on reaction with hot and concentrated sodium hydroxide gives: [Main Jan. 12, 2019 (II)] (a) \(\mathrm{Cl}\) and \(\mathrm{ClO}_{3}^{-}\) (b) \(\mathrm{Cl}^{-}\)and \(\mathrm{ClO}^{-}\) (c) \(\mathrm{ClO}_{3}^{-}\)and \(\mathrm{ClO}_{2}^{-}\) (d) \(\mathrm{Cl}^{-}\)and \(\mathrm{ClO}_{2}^{-}\)

White P reacts with caustic soda. The products are \(\mathrm{PH}_{3}\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{2}\). This reaction is an example of \([1980]\) (a) Oxidation (b) Reduction (c) oxidation and reduction (d) Neutralisation

Which of the following compound has a P-P bond? [Main Online April 11, 2015] (a) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}\) (b) \(\left(\mathrm{HPO}_{3}\right)_{3}\) (c) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) (d) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\)
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