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 \(0.10 \mathrm{M}\) solution of benzoic acid or a \(0.10 \mathrm{M}\) solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} & \mathrm{NH}_{2}(a q) \rightleftharpoons \\ & \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}+(a q) \end{aligned} $$

Short answer

\(a)\) The conjugate base of benzoic acid is \( C_6H_5COO^-\) and the conjugate acid of aniline is \( C_6H_5NH_3^+\). \(b)\) A \(0.10 \mathrm{M}\) solution of anilinium chloride is more acidic than a \(0.10 \mathrm{M}\) solution of benzoic acid. \(c)\) The value of the equilibrium constant for the given reaction is \(K = \frac{K_a}{K_b}\), where \(K_a\) is for benzoic acid and \(K_b\) is for anilinium ion.

Step-by-step solution

a) Conjugate base of benzoic acid and conjugate acid of aniline

To find the conjugate base of benzoic acid, we look at what happens when it loses an H+ ion: \( C_6H_5COOH \rightarrow C_6H_5COO^- + H^+ \) The conjugate base of benzoic acid is \( C_6H_5COO^-\). To find the conjugate acid of aniline, we look at what happens when it gains an H+ ion: \( C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+ \) The conjugate acid of aniline is \( C_6H_5NH_3^+\).

b) Comparing the acidity of 0.10 M benzoic acid and 0.10 M anilinium chloride

Since benzoic acid is a weak acid, we can use the given Ka value to determine the concentration of H+ for a 0.10 M solution: \( K_a = \frac{[H^+][C_6H_5COO^-]}{[C_6H_5COOH]} \) Anilinium chloride is a strong electrolyte and dissociates completely into anilinium ions and chloride ions. Since anilinium ion is the conjugate acid of aniline, we can use the given Ka value of aniline to find the Kb value for anilinium ion: \( K_w = K_a \times K_b \) Then, we can determine the concentration of H+ for the 0.10 M anilinium ion solution using the calculated Kb value: \( K_b = \frac{[H^+][C_6H_5NH_2]}{[C_6H_5NH_3^+]} \) By comparing the concentration of H+ ions for each solution, we can determine which solution is more acidic.

c) Equilibrium constant for the given reaction

To find the equilibrium constant for the given reaction, we can use the formula K = K1/K2, where K1 is the Kа for the weak acid, and K2 is the Kb for the weak base conjugate: \(K = \frac {[C_6H_5COO^-][C_6H_5NH_3^+]}{[C_6H_5COOH][C_6H_5NH_2]}\) \(K = \frac{K_a}{K_b};\) Use the given \(K_a\) for benzoic acid and the calculated \(\mathrm{K}_b\) for anilinium to find the equilibrium constant (K) for the given reaction.

Key concepts

Benzoic Acid

Benzoic acid, denoted as \( C_6H_5COOH \), is a simple aromatic carboxylic acid that is used variously in the food industry as a preservative and in the creation of various organic compounds. It is weakly acidic, meaning that in an aqueous solution, it partially dissociates to produce \( H^+ \) ions and the conjugate base \( C_6H_5COO^- \).

The strength of its acidity can be quantified by an equilibrium constant, called the acid dissociation constant (Ka). For benzoic acid, \( K_{a}=6.3 \times 10^{-5} \), indicating that the compound does not completely ionize in water. When solving problems involving benzoic acid, recognizing its weak acid behavior is crucial for predicting its reactions with bases, its dissociation in solutions, and the resulting pH.

Aniline

Aniline, with the chemical formula \( C_6H_5NH_2 \), is a primary amine and a derivative of benzene. It acts as a weak base in reactions. Aniline accepts protons in solution, forming its conjugate acid \( C_6H_5NH_3^+ \).

The acidity constant \( K_{a} \) for aniline is considerably smaller than that of benzoic acid, sitting at \( K_{a}=4.3 \times 10^{-10} \), which reflects the tendency of aniline to behave as a base rather than an acid. Consequently, aniline is more prone to accept a proton rather than donate one. This property is critical when predicting the behavior of aniline in acid-base reactions and when assessing its role in creating compounds such as anilinium chloride, a derivative used in the question prompt.

Conjugate Acid-Base Pairs

The concept of conjugate acid-base pairs is fundamental to understanding acid-base reactions. A conjugate base is the species that remains after an acid has donated a proton, whereas a conjugate acid is the species that forms when a base gains a proton. For instance, when benzoic acid loses a proton, it forms the conjugate base \( C_6H_5COO^- \). On the other hand, aniline becomes a conjugate acid, \( C_6H_5NH_3^+ \), when it accepts a proton.

These transformations are reversible and are at the heart of the acid-base equilibrium. Every acid has its counterpart base, and every base has a corresponding acid, forming a conjugate pair. This pairing allows the prediction of the direction and extent of chemical reactions in aqueous solutions, depending on the strength of the acids and bases involved.

Equilibrium Constant

The equilibrium constant (K) is critical in quantifying the position of equilibrium in a chemical reaction. For acid-base equilibria, the constants \( K_a \) and \( K_b \) tell us how much the acid or base dissociates in water. The weaker the acid or base, the smaller the value of its \( K_a \) or \( K_b \) respectively. Conversely, strong acids and bases have larger constants and disassociate almost completely in solution.

For the given reaction between benzoic acid and aniline, the equilibrium constant \( K \) can be derived from the \( K_a \) of benzoic acid and the \( K_b \) of the aniline's conjugate acid, which can be calculated from \( K_w \) (the ion-product constant of water), since \( K_w = K_a \times K_b \). Knowing \( K \) can help us predict the composition of the equilibrium mixture, thus informing us about the relative concentrations of reactants and products at equilibrium.