Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

When brine solution is saturated with \(\mathrm{NH}_{3}\) and \(\mathrm{CO}_{2}\) a slightly alkaline white sodium salt (A) is formed which has \(\mathrm{pH}\) of nearly \(8.4 .\) (A) on heating liberates a gas (B) leaving a highly alkaline residue (C) of \(\mathrm{pH}\) nearly 10 to 11 . Gas (B) is colourless and turns a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) milky. Identify (B). (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (b) \(\mathrm{NaHCO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{~S}\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Gas (B) is \(\mathrm{CO}_2\), making option (b) \(\mathrm{NaHCO}_3\) correct.

Step 1: Analyze the Saturated Brine Solution

When a brine solution saturated with ammonia \(\mathrm{NH}_3\)\u00a0and carbon dioxide \(\mathrm{CO}_2\) is formed, a white sodium salt (\(\mathrm{A}\)) is identified which has a \(\text{pH}\) of roughly 8.4. The slightly alkaline nature of \(\text{pH}~8.4\) indicates the formation of \(\mathrm{NaHCO}_3\) or sodium bicarbonate.

Step 2: Heating and Decomposition

When \(\mathrm{NaHCO}_3\) is heated, it decomposes to form a gas \(\mathrm{B}\) which is \(\mathrm{CO}_2\), and it leaves behind a highly alkaline residue \(\mathrm{C}\) of \(\mathrm{Na}_2\mathrm{CO}_3\) which has a \(\text{pH}\) between 10 and 11.

Step 3: Gas Identification

The colorless gas \(\mathrm{B}\) that turns lime water, \(\mathrm{Ca(OH)}_2\), milky is identified as \(\mathrm{CO}_2\), because when carbon dioxide reacts with lime water, it forms \(\mathrm{CaCO}_3\), which is insoluble and makes the solution appear cloudy or milky.

Key Concepts

Brine Solution Chemistry

Brine solution chemistry involves the reaction of a saltwater solution with other chemical substances to form new compounds. In this case, the brine solution is saturated with ammonia (\(\text{NH}_3\)) and carbon dioxide (\(\text{CO}_2\)) to produce a sodium compound known as sodium bicarbonate (\(\text{NaHCO}_3\)). When brine, which is essentially a concentrated solution of sodium chloride (\(\text{NaCl}\)), interacts with ammonia and carbon dioxide, certain reactions occur that lead to the precipitation of this slightly alkaline white solid.

• Sodium Bicarbonate Formation: The ammonia and carbon dioxide dissolve in the brine to produce sodium bicarbonate, indicated by a white precipitate.
• Chemical Process: The combination can be written as: \[ \text{NH}_3 + \text{CO}_2 + \text{NaCl} + \text{H}_2\text{O} \rightarrow \text{NaHCO}_3 + \text{NH}_4\text{Cl} \]
• pH Level: The resulting solution where sodium bicarbonate is formed has a slightly alkaline pH of approximately 8.4.

Brine solutions are not only interesting for the salts they help to produce but also for understanding the complex interactions between solvents and various solutes in aqueous solutions. These reactions underscore how different ions and molecules interact in water, facilitating the formation of products with varying properties and chemical behaviors.

pH and Alkalinity

The concepts of pH and alkalinity are critical in understanding how substances interact in solution, especially in brine solutions where sodium salts are present. pH is a measure of how acidic or basic a solution is, on a scale of 0 to 14. Alkalinity, on the other hand, refers to the ability of a solution to neutralize acids. In the context of sodium bicarbonate (\(\text{NaHCO}_3\)) reactions:

• pH Values: Sodium bicarbonate solutions have a pH of about 8.4, indicating that they are mildly alkaline, giving a clue to its basic character.
• Change in Alkalinity: When sodium bicarbonate (\(\text{NaHCO}_3\)) is heated and decomposes into sodium carbonate (\(\text{Na}_2\text{CO}_3\)), the pH increases to about 10 - 11. This shows a significant rise in alkalinity since sodium carbonate is a stronger base.
• Why it Matters: Understanding pH and alkalinity changes informs chemists about the behavior of compounds during reactions and in different conditions, making this basic chemistry crucial for practical applications like neutralizing acids.

These properties not only affect chemical processes and stability in reactions but also have broad applications—from cooking and cleaning to environmental management and industrial manufacturing.

Gas Identification in Reactions

Gas identification in chemical reactions is an integral part of understanding how substances transform and interact. When sodium bicarbonate (\(\text{NaHCO}_3\)) is decomposed by heating, carbon dioxide (\(\text{CO}_2\)) is released as a gas:

• Carbon Dioxide Release: Upon heating, sodium bicarbonate decomposes into sodium carbonate (\(\text{Na}_2\text{CO}_3\)), water vapor, and carbon dioxide. This process can be described by the reaction: \[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]
• Characterizing Carbon Dioxide: The colorless gas (\(\text{CO}_2\)) is identified by its reaction with calcium hydroxide (\(\text{Ca(OH)}_2\)), turning limewater milky. This reaction produces calcium carbonate (\(\text{CaCO}_3\)), which is visible as a milky precipitate: \[ \text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \]
• Significance: Identifying gases like carbon dioxide is crucial in various scientific disciplines, as it can help monitor reactions, confirm experimental results, and ensure safety during processes that involve gas production.

Gas evolution and identification are not just textbook chemistry concepts—they actively play roles in understanding industrial processes, household tasks, and even climate studies.