Problem 43 - Chemistry and Chemical Reactivity

Question

Complete and balance the following equations: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\ell)+\mathrm{HCl}(\mathrm{aq}) \rightarrow\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow\)

Step-by-Step Solution

Answer

(a) \(\text{C}_{6}\text{H}_{5}\text{NH}_{2} + \text{HCl} \rightarrow \text{C}_{6}\text{H}_{5}\text{NH}_{3}^+ \text{Cl}^- \); (b) \(2(\text{CH}_{3})_{3}\text{N} + \text{H}_{2}\text{SO}_{4} \rightarrow 2((\text{CH}_{3})_{3}\text{NH})^{+} (\text{SO}_{4})^{2-} \).

Step 1: Identify Reactants and Products for Equation (a)

The reactants for equation (a) are aniline ( \(\text{C}_{6}\text{H}_{5}\text{NH}_{2}(\ell)\) ) and hydrochloric acid ( \(\text{HCl}(\text{aq})\) ). The product formed is anilinium chloride, \(\text{C}_{6}\text{H}_{5}\text{NH}_{3}^+\text{Cl}^-\text{(aq)}\) . Aniline reacts with HCl to gain a proton and form the anilinium ion.

Step 2: Write the Balanced Equation for (a)

To write the balanced chemical equation, ensure that the number and type of atoms on the reactant side is equal to the number of same atoms on the product side. \( \text{C}_{6}\text{H}_{5}\text{NH}_{2}(\ell) + \text{HCl}(\text{aq}) \rightarrow \text{C}_{6}\text{H}_{5}\text{NH}_{3}^+ \text{Cl}^- (\text{aq}) \)

Step 3: Identify Reactants and Products for Equation (b)

The reactants for equation (b) are trimethylamine ( \((\text{CH}_{3})_{3}\text{N}(\text{aq})\) ) and sulfuric acid ( \(\text{H}_{2}\text{SO}_{4}(\text{aq})\) ). The products formed are trimethylammonium sulfate, \(((\text{CH}_{3})_{3}\text{NH})(\text{SO}_{4})_{0.5}^-(\text{aq})\) .

Step 4: Write the Balanced Equation for (b)

To ensure the equation is balanced, match the number of each type of atom on each side of the equation. The balanced equation is: \( 2(\text{CH}_{3})_{3}\text{N}(\text{aq}) + \text{H}_{2}\text{SO}_{4}(\text{aq}) \rightarrow 2((\text{CH}_{3})_{3}\text{NH})^{+}(\text{aq}) (\text{SO}_{4})^{2-}(\text{aq}) \)

Key Concepts

Aniline Hydrochloride Formation

Aniline hydrochloride forms when aniline, a simple aromatic amine, reacts with hydrochloric acid. This process is a classic example of acid-base chemistry where the basic amino group in aniline accepts a proton from the hydrochloric acid.

This leads to the formation of the anilinium ion. The overall reaction can be summarized as follows:

• The reactants are aniline (\(C_{6}H_{5}NH_{2}\)) and hydrochloric acid (\(HCl\)).
• The reaction results in the formation of anilinium chloride (\(C_{6}H_{5}NH_{3}^{+}Cl^{-}\)).

The balanced chemical equation is:

\[ C_{6}H_{5}NH_{2}( ext{l}) + HCl( ext{aq}) \rightarrow C_{6}H_{5}NH_{3}^{+}Cl^{-}( ext{aq}) \]

This equation shows the conversion of a neutral aniline molecule into a charged anilinium ion in the presence of hydrochloric acid.

Trimethylamine Reaction

Trimethylamine reacts with sulfuric acid to form trimethylammonium sulfate. This reaction involves neutralization, where the trimethylamine base accepts protons from sulfuric acid.

• The reactants are trimethylamine \((CH_{3})_{3}N\) and sulfuric acid \(H_{2}SO_{4}\).
• The products include trimethylammonium sulfate \(((CH_{3})_{3}NH_{2})^{+}(SO_{4})^{2-}\).

The balanced reaction equation is:

\[ 2(CH_{3})_{3}N( ext{aq}) + H_{2}SO_{4}( ext{aq}) \rightarrow 2((CH_{3})_{3}NH)^{+} ( ext{aq}) (SO_{4})^{2-}( ext{aq}) \]

This equation indicates that two molecules of trimethylamine are needed to interact with one molecule of sulfuric acid.

Anilinium Ion

The anilinium ion is the result of aniline accepting a proton from hydrochloric acid. In this process, the nitrogen atom in aniline gains a hydrogen ion (proton) because it acts as a Lewis base.

Understanding the anilinium ion is essential because:

• It carries a positive charge \(NH_{3}^{+}\), making it ionic.
• Ionic forms are more water-soluble, important in many chemical applications.

The creation of the anilinium ion illustrates how electron-rich nitrogen can stabilize additional positive charge through protonation.

Trimethylammonium Sulfate

Trimethylammonium sulfate forms when trimethylamine reacts with sulfuric acid, and it involves two key steps: protonation and balancing.

Trimethylamine, a strong base, accepts protons from sulfuric acid:

• The resulting trimethylammonium ion \(((CH_{3})_{3}NH)^{+}\) is paired with sulfate ions \((SO_{4})^{2-}\).

The process demonstrates a typical acid-base reaction resulting in a salt:

• This salt is highly soluble in water due to its ionic nature.
• The balanced reaction highlights the need for two trimethylammonium ions to pair with sulfate's double negative charge.

Trimethylammonium sulfate is thus effectively formed by the donation of protons from the acid to the amine base, creating a stable ionic compound.