Problem 8 - General Chemistry

Question

Write the equation for the ionization of aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\) in aqueous solution. Write the expression for \(K_{b}\)

Step-by-Step Solution

Answer

The equation is \( \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+ + \mathrm{OH}^- \), and \( K_b = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+][\mathrm{OH}^-]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}]} \).

Step 1: Understanding Ionization of Aniline

Aniline, \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}\), is a weak base. When it ionizes in water, it accepts a proton from water (acting as the Bronsted-Lowry base) to form its conjugate acid, anilinium ion \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+\), and hydroxide ion \(\mathrm{OH}^-\).

Step 2: Writing the Ionization Reaction

The ionization reaction of aniline in water can be written as follows:\[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2 (aq)} + \mathrm{H}_{2}\mathrm{O}_{(l)} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+_{(aq)} + \mathrm{OH}^-_{(aq)}\]

Step 3: Writing the Expression for the Base Ionization Constant \(K_b\)

The expression for the base ionization constant \(K_b\) follows the law of mass action. For the equilibrium reaction, \(K_b\) is given by:\[K_b = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+][\mathrm{OH}^-]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}]}\]where the square brackets indicate the concentration of each species at equilibrium.

Key Concepts

Weak Bases

A weak base is a substance that partially ionizes in water. Unlike strong bases, which completely dissociate, weak bases only dissociate to a limited extent in aqueous solutions. Aniline, with the chemical formula \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}\), is a classic example of a weak base. When dissolved in water, it does not fully convert into its ionized form. This partial ionization is crucial in determining the acidic or basic strength of the solution.

• Aniline partially ionizes, accepting protons from water and leading to only a small concentration of hydroxide ions \(\mathrm{OH}^-\).
• This makes the solution alkaline but not as strongly basic as if a strong base were dissolved.

Bronsted-Lowry Base

The Bronsted-Lowry theory provides a broad definition of acids and bases. According to this theory, a base is a proton (\(\mathrm{H}^+\)) acceptor. This is different from the classical definition that strictly considers bases that release \(\mathrm{OH}^-\) ions. In the case of aniline, it acts as a Bronsted-Lowry base when it accepts a proton from water. This forms the anilinium ion \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+\) and rudimentary hydroxide ions.

• The reaction of aniline with water is an example where aniline acts as a Bronsted-Lowry base.
• It illustrates the concept that base strength is not merely the ability to produce \(\mathrm{OH}^-\) ions but also its ability to accept protons.

Base Ionization Constant

The base ionization constant, represented as \(K_b\), is a measure of the strength of a base in a solution. It quantifies the extent to which a base ionizes in water, providing insight into its proton accepting ability. For weak bases like aniline, \(K_b\) is typically less than one, indicating limited ionization.

• For aniline, the \(K_b\) expression is given as: \[K_b = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+][\mathrm{OH}^-]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}]}\]
• A higher \(K_b\) value corresponds to a stronger base since it suggests a greater concentration of ions at equilibrium.

Equilibrium Reaction

In chemistry, equilibrium occurs when the rates of the forward and reverse reactions are equal. This means the concentration of reactants and products remains constant over time. For aniline in water, this dynamic balance involves the constant switching between the reactants \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}\) and water, and the products \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+\) and \(\mathrm{OH}^-\).

• This equilibrium is crucial for calculating \(K_b\) and understanding the properties of weak bases like aniline.
• The double arrow \(\rightleftharpoons\) used in the chemical equation represents the reversible nature typical of equilibrium reactions.

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It depicts the conversion of reactants to products. The ionization of aniline is described by the following chemical equation:
\[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2 (aq)} + \mathrm{H}_{2}\mathrm{O}_{(l)} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^+_{(aq)} + \mathrm{OH}^-_{(aq)}\]
This equation indicates that aniline and water participate in the reaction to form anilinium ions and hydroxide ions.

• The state's notation, such as \((aq)\) for aqueous, conveys the physical state of the substances involved.
• Balance is key in chemical equations, ensuring equal atoms on both sides of the equation.
• For weak bases like aniline, partial dissociation is depicted by the use of equilibrium arrows.