Question

Which compound will liberate \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\) solution? (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\) (d) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{OH}^{-}\)

Short answer

(c) \(\mathrm{CH}_{3} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\) liberates \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\).

Step-by-step solution

Understanding the Equation

The liberation of \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\) occurs when an acid reacts with bicarbonate. This can be represented by the equation: \(\mathrm{NaHCO}_{3} + \text{Acid} \rightarrow \mathrm{Na}^{+} + \mathrm{CO}_{2} + \text{Water + Salt}\). We need to identify which compound among the options acts as an acid in solution.

Analyze Compound (a)

\(\mathrm{CH}_{3}\mathrm{NH}_{2}\) is methylamine, which is a weak base. It does not act as an acid and will not liberate \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\).

Analyze Compound (b)

\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\) is acetamide, which does not dissociate in a way to provide protons as required by an acid. It cannot liberate \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\).

Analyze Compound (c)

\(\mathrm{CH}_{3}\mathrm{~N}^{+}\mathrm{H}_{3}\mathrm{Cl}^{-}\) is ammonium chloride. The \(\mathrm{NH}_{4}^{+}\) ion acts as a weak acid, which can donate protons to react with \(\mathrm{NaHCO}_{3}\), liberating \(\mathrm{CO}_{2}\).

Analyze Compound (d)

\(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{OH}^{-}\) is tetramethylammonium hydroxide, which is a strong base. It will not behave as an acid and does not liberate \(\mathrm{CO}_{2}\) from \(\mathrm{NaHCO}_{3}\).

Conclusion

Among the given options, only \(\mathrm{CH}_{3} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\), which contains \(\mathrm{NH}_{4}^{+}\), can give protons to \(\mathrm{NaHCO}_{3}\), leading to \(\mathrm{CO}_{2}\) liberation.

Key concepts

Bicarbonate Reactions

Bicarbonate reactions involve interactions between bicarbonate ions, like those found in sodium bicarbonate (\(\text{NaHCO}_3\)), and acidic or basic substances. These reactions are essential in both natural and industrial processes.
Bicarbonate itself is an important buffer, capable of neutralizing acids. When bicarbonate reacts with an acid, it undergoes a decomposition reaction, liberating carbon dioxide (\(\text{CO}_2\)).

• The general reaction is: \[\text{HCO}_3^- + \text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]

Understanding which compounds can act as acids is crucial in identifying which reactions will occur. For instance, weak acids can effectively react with bicarbonate and achieve this decomposition, whereas bases cannot initiate this process.

Compound Analysis

When analyzing compounds to determine their behavior in chemical reactions, we typically look at their ability to donate or accept protons. In chemistry, an acid is a proton donor, while a base is a proton acceptor.
To evaluate whether a compound can facilitate the reaction described in a bicarbonate reaction, it needs to function as an acid in the presence of bicarbonate.

• For example, methylamine (\(\text{CH}_3\text{NH}_2\)) is a weak base and is unable to provide protons, resulting in no carbon dioxide liberation.


• Acetamide \(\text{CH}_3\text{CONH}_2\) does not readily dissociate to donate protons and therefore does not function as an acid in these conditions.

In contrast, ammonium chloride \(\text{CH}_3\text{~N}^{+}\text{H}_3\text{Cl}^{-}\) contains \(\text{NH}_4^+\) ions which serve as weak acids capable of donating protons to the bicarbonate, resulting in the release of carbon dioxide.

Ammonium Chloride Acidic Behavior

Ammonium chloride (\(\text{CH}_3\text{~N}^{+}\text{H}_3\text{Cl}^{-}\)) exhibits acidic properties primarily due to the presence of the ammonium ion (\(\text{NH}_4^+\)). This ion is capable of acting as a weak acid by donating a proton to a base or, in this context, a bicarbonate ion.
When \(\text{NH}_4^+\) donates a proton, it forms ammonia (\(\text{NH}_3\)) and helps release \(\text{CO}_2\) from the bicarbonate:

• The reaction can be represented as:\[\text{NH}_4^+ + \text{HCO}_3^- \rightarrow \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O}\]

This ability of ammonium chloride to liberate \(\text{CO}_2\) makes it distinct from other compounds which cannot donate protons and hence, not act as acids in similar reactions.