Question

Benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) are both derivatives of benzene. Benzoic acid is an acid with \(K_{a}=6.3 \times 10^{-5}\) and aniline is a base with \(K_{a}=4.3 \times 10^{-10} .\) (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\) is a strong electrolyte that dissociates into anilinium ions \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right)\) and chloride ions. Which will be more acidic, a 0.10\(M\) solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$\begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{5} \operatorname{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q) \rightleftharpoons} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(a q)}\end{array}$$

Short answer

(a) The conjugate base of benzoic acid is \(C_6H_5COO^-(aq)\), and the conjugate acid of aniline is \(C_6H_5NH_3^+(aq)\). (b) A 0.10 M solution of benzoic acid is more acidic with a pH of 2.60 compared to a 0.10 M solution of anilinium chloride with a pH of 2.82. (c) The equilibrium constant, K, for the given reaction is approximately \(4.29\times10^{-10}\).

Step-by-step solution

Identify the conjugate pairs in the given acid and base

Write down the chemical reactions for benzoic acid and aniline ionization: Benzoic acid ionization: \(C_6H_5COOH(aq) \rightleftharpoons C_6H_5COO^-(aq) + H^+(aq)\) Aniline ionization: \(C_6H_5NH_2(aq) + H^+(aq) \rightleftharpoons C_6H_5NH_3^+(aq) \)

Determine conjugate base and conjugate acid

From the ionization reactions: Conjugate base of benzoic acid: \(C_6H_5COO^-(aq)\) Conjugate acid of aniline: \(C_6H_5NH_3^+(aq)\) #b) Comparison of the acidity of benzoic acid and anilinium chloride solutions.#

Calculate pH of 0.10 M benzoic acid solution

We use the acid ionization constant expression for benzoic acid: \(K_{a}=\frac{[H^+][C_6H_5COO^-]}{[C_6H_5COOH]}\) Let the hydrogen ion concentration be x: \(K_{a} = \frac{x\cdot x}{0.10 - x} = 6.3\times10^{-5}\) Solve the equation for x(use the approximation, x<<0.10): x = 0.00251 pH of benzoic acid solution = -log(0.00251) ≈ 2.60

Calculate pH of 0.10 M anilinium chloride solution

Anilinium ion in the solution reacts with water: \(C_6H_5NH_3^+(aq) + H_2O(l) \rightleftharpoons C_6H_5NH_2(aq) + H_3O^+(aq)\) Using the aniline's ionization constant: \(K_{a}=4.3 \times 10^{-10}\) Find Kb for aniline: \(K_w=K_a\cdot K_b\) => \(K_b= \frac{1\times10^{-14}}{4.3 \times 10^{-10}} = 2.33 \times 10^{-5}\) Equilibrium equation for anilinium ion reaction: \(K_b = \frac{[C_6H_5NH_2][H_3O^+]}{[C_6H_5NH_3^+]}\) Let the hydrogen ion concentration be y: \(K_b = \frac{y\cdot y}{0.10 - y} = 2.33\times10^{-5}\) Solve the equation for y (use the approximation, y<<0.10): y = 0.00153 pH of anilinium chloride solution = -log(0.00153) ≈ 2.82

Compare the pH values of the solutions

pH of benzoic acid solution ≈ 2.60 pH of anilinium chloride solution ≈ 2.82 Since the pH of the benzoic acid solution is lower, the 0.10 M benzoic acid solution is more acidic than the 0.10 M anilinium chloride solution. #c) Determining the equilibrium constant.#

Write the ionization equations for the reaction

Ionization equation for benzoic acid: \(C_6H_5COOH(aq) \rightleftharpoons C_6H_5COO^-(aq) + H^+(aq)\) Ionization equation for aniline: \(C_6H_5NH_2(aq) + H^+(aq) \rightleftharpoons C_6H_5NH_3^+(aq) \) Equilibrium equation: \(C_6H_5COOH(aq) + C_6H_5NH_2(aq) \rightleftharpoons C_6H_5COO^-(aq) + C_6H_5NH_3^+(aq)\)

Use ionization constants to find the equilibrium constant

Using the equilibrium constants for benzoic acid and aniline, as well as their ionization equations and the equilibrium equation, the equilibrium constant K can be derived as: \(K=\frac{K_{a}[H^+]}{K_b}\) With \(K_{a}=6.3\times10^{-5}\), \(K_b=2.33\times10^{-5}\), and \([H^+]=10^{-14}/K_a\), we substitute the values into the equation and solve for K: \(K=\frac{6.3\times 10^{-5}\times(10^{-14}/(6.3\times10^{-5}))}{2.33\times10^{-5}}\) \(K=\frac{10^{-14}}{2.33\times10^{-5}} \) \(K=4.29\times10^{-10}\) Thus, the value of the equilibrium constant K for the given reaction is about 4.29 × 10⁻¹⁰.

Key concepts

Benzoic Acid Conjugate Base

Understanding the conjugate base of benzoic acid is key when studying acid-base chemistry. Benzoic acid \(C_6H_5COOH\) ionizes in water to form its conjugate base, benzoate \(C_6H_5COO^-\), and a proton \(H^+\). This reaction can be written as \[C_6H_5COOH(aq) \rightleftharpoons C_6H_5COO^-(aq) + H^+(aq)\]. The benzoate ion is the species that remains after benzoic acid has donated a proton.

The strength of an acid is often compared by looking at its acid ionization constant \(K_a\). For benzoic acid, \(K_a = 6.3 \times 10^{-5}\), indicating it's a weak acid. In aqueous solution, benzoic acid only partially dissociates, establishing an equilibrium between the acid and its conjugate base.

When solving chemical equilibrium problems, recognizing conjugate acid-base pairs is crucial. The ability of the benzoate ion to potentially accept a proton makes it an important species in buffer solutions and other chemical reactions.

Aniline Conjugate Acid

The conjugate acid of aniline \(C_6H_5NH_2\) is anilinium ion \(C_6H_5NH_3^+\). Aniline is a weak base with its own acid dissociation constant \(K_a = 4.3 \times 10^{-10}\). It accepts a proton from water or an acid to form its conjugate acid, which is represented by the equation \[C_6H_5NH_2(aq) + H^+(aq) \rightleftharpoons C_6H_5NH_3^+(aq)\].

In this reaction, aniline acts as a proton acceptor, which is the basic principle defining a Bronsted-Lowry base. The degree to which this reaction occurs depends on the strength of the base, governed by \(K_b\), the base dissociation constant. Since \(K_w=K_a\cdot K_b\), you can deduce \(K_b\) from known values, allowing calculation of equilibria in solutions containing aniline or its conjugate acid.

pH Calculation

pH calculation is a process to determine the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with 7 being neutral. Values below 7 indicate acidity, and values above 7 indicate basicity. pH is defined as the negative logarithm of the hydrogen ion concentration \(pH = -\log[H^+]\).

For acids and bases like benzoic acid and anilinium chloride, we often need to make an approximation for simplification, assuming that the concentration of the acid or base doesn’t change significantly when calculating \( [H^+] \). Once \( [H^+] \), or the concentration of hydrogen ions, is known, the pH can be calculated.

For example, in a 0.10 M benzoic acid solution, the calculation implies a pH of approximately 2.60, indicating a fairly acidic solution. This knowledge aids in understanding reaction conditions and the behavior of substances in different environments.

Equilibrium Constant

The equilibrium constant \(K\) is a value that expresses the ratio of concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. For acid-base reactions, we often refer to \(K_a\), the acid dissociation constant, or \(K_b\), the base dissociation constant.

In the exercise, the equilibrium involves the acid-base reaction between benzoic acid and aniline. To find the overall equilibrium constant, we derive it from \(K_a\) and \(K_b\) of the reactants. The final expression for the equilibrium constant \(K\) takes the form \[K=\frac{[C_6H_5COO^-][C_6H_5NH_3^+]}{[C_6H_5COOH][C_6H_5NH_2]}\], combining individual ionization processes into one equation.

The value of \(K\) gives us insight into the position of equilibrium. A low \(K\) suggests that the reactants are favored, while a high \(K\) indicates a tendency to form products. These principles are integral to understanding and predicting the outcome of chemical reactions.