Problem 105 - Chemistry The Central Science In Si Units

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 \(M\) solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium?

Step-by-Step Solution

Answer

(a) Conjugate base: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\), conjugate acid: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). (b) Benzoic acid is more acidic. (c) \(K_{eq} \approx 2.70\).

Step 1: Determine Conjugate Base and Acid

For (a), to find the conjugate base of benzoic acid (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\)), we remove a proton (H\(^+\)) which results in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\). To find the conjugate acid of aniline (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\)), we add a proton, resulting in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\).

Step 2: Compare Acidic Strengths

For (b), compare the acidity of benzoic acid and anilinium ion. Benzoic acid's acidity can be described by its \(K_a\) value of \(6.3 \times 10^{-5}\). The anilinium ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\), is the conjugate acid of a weak base (aniline), with a \(K_a\) from its \(K_b\). Calculate the \(K_a\) of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) using \(K_w = 1.0 \times 10^{-14}\) and \(K_b\) of aniline (\(4.3 \times 10^{-10}\)): \(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}} \approx 2.33 \times 10^{-5}\). Since benzoic acid's \(K_a = 6.3 \times 10^{-5}\) is greater than \(2.33 \times 10^{-5}\), benzoic acid is the stronger acid.

Step 3: Find Equilibrium Constant for Reaction

For (c), the equilibrium involves the reaction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). The equilibrium constant \(K_{eq}\) can be found using the relationship: \(K_{eq} = \frac{K_a(\text{acid 1})}{K_a(\text{acid 2})}\). \(K_a\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) is \(6.3 \times 10^{-5}\) and for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) is calculated as \(2.33 \times 10^{-5}\). Thus, \(K_{eq} = \frac{6.3 \times 10^{-5}}{2.33 \times 10^{-5}} \approx 2.70\).

Key Concepts

The Role of Conjugate Acid-Base Pairs in Equilibrium

Conjugate acid-base pairs are a fundamental concept in understanding acid-base reactions and equilibrium. In simple terms, a conjugate acid is what you get when a base gains a proton (H\(^+\)), and a conjugate base is what remains when an acid loses a proton.
For instance, in the case of benzoic acid (\(\text{C}_6\text{H}_5\text{COOH}\)), when it loses a proton, the result is its conjugate base, benzoate ion (\(\text{C}_6\text{H}_5\text{COO}^-\)). Similarly, aniline (\(\text{C}_6\text{H}_5\text{NH}_2\)) gains a proton to form its conjugate acid, anilinium ion (\(\text{C}_6\text{H}_5\text{NH}_3^+\)).
These transformations illustrate how acids and bases can transform into their conjugate pairs, allowing reactions to achieve equilibrium. Understanding these pairs helps predict the direction of reactions and the strength of the resulting solutions.

Understanding the Acid Dissociation Constant (Ka)

The acid dissociation constant, denoted as \(K_a\), is a numerical value that expresses the strength of an acid in a solution. It is an equilibrium constant that measures how well an acid can donate a proton to water and become its conjugate base.
Formally, for a generic acid (\(HA\)), the expression for \(K_a\) is given by: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] where \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid at equilibrium.
In the exercise, benzoic acid has a \(K_a\) of \(6.3 \times 10^{-5}\), showing its relative strength to donate protons. Contrastingly, anilinium ion, the conjugate acid of aniline, has a \(K_a\) calculated as \(2.33 \times 10^{-5}\). The larger the \(K_a\), the stronger the acid, meaning it dissociates more in water. Hence, comparing \(K_a\) values helps predict which solution is more acidic.

Equilibrium Constant (Keq) and Its Importance in Reactions

The equilibrium constant, represented as \(K_{eq}\), is essential for understanding how far a reaction proceeds before reaching equilibrium. It provides insight into the ratio of products to reactants at equilibrium for a given reaction.
In the equilibrium reaction of benzoic acid with aniline forming benzoate ions and anilinium ions, the \(K_{eq}\) can be found using the ratio of \(K_a\) values: \[ K_{eq} = \frac{K_a(\text{benzoic acid})}{K_a(\text{anilinium ion})} \] Substituting the given values, \(6.3 \times 10^{-5}\) for benzoic acid and \(2.33 \times 10^{-5}\) for anilinium ion, gives \(K_{eq} \approx 2.70\).
This value indicates that the reaction favors product formation (benzoate ions and anilinium ions) over reactants (benzoic acid and aniline). Understanding \(K_{eq}\) is vital for predicting the position of equilibrium in reactions.