Problem 105 - Chemistry: The Central Science in SI Units, Global Edition

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) The conjugate base of benzoic acid is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\), and the conjugate acid of aniline is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). b) The \(0.10 \mathrm{M}\) solution of anilinium chloride will be more acidic than the \(0.10 \mathrm{M}\) solution of benzoic acid. c) The equilibrium constant for the given equilibrium is \(K = \frac{K_{b} \times K_{w}}{K_{a}}= 6.3 \times 10^{-10}\).

Step 1: 1. Determine the conjugate base of benzoic acid and the conjugate acid of aniline.

To find the conjugate base of benzoic acid, we will remove one proton (H+) from the acid. For aniline, we will add one proton (H+) to the base to find the conjugate acid. a) Conjugate base of benzoic acid: Remove one proton (H+) from \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\). We have the conjugate base as \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\). Conjugate acid of aniline: Add one proton (H+) to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). We have the conjugate acid as \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\).

Step 2: 2. Compare the acidity of 0.10 M benzoic acid solution and 0.10 M anilinium chloride solution.

To compare the acidity of the benzoic acid solution and the anilinium chloride solution, we can calculate the \(\mathrm{pH}\) of each solution. b) For the benzoic acid solution: Using the given \(K_{a}=6.3 \times 10^{-5}\) for benzoic acid, we can apply the formula for \(\mathrm{pH}\). \[ \mathrm{pH} = -\log(\sqrt{(0.10) \times (6.3 \times 10^{-5})})\] Calculate the pH of benzoic acid solution. For the anilinium chloride solution: Given that anilinium chloride is a strong electrolyte, it dissociates completely into anilinium ions and chloride ions. This means that the concentration of anilinium ions will be 0.10 M. Now we can use the Kb value of aniline \(K_{b}= \frac{K_w}{K_a}= \frac{1 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5}\). To calculate \(\mathrm{pH}\) of the anilinium chloride solution, we can use the formula for \(\mathrm{pOH}\) and \(\mathrm{pH} = 14 - \mathrm{pOH}\): \[ \mathrm{pOH} = -\log(\sqrt{(0.10) \times (2.3 \times 10^{-5})})\] Calculate the pH of the anilinium chloride solution. Compare the two pH values to determine which solution is more acidic.

Step 3: 3. Find the equilibrium constant for the given equilibrium.

Recall that for the reaction: $$\mathrm{HA} \rightleftharpoons \mathrm{A}^{-} + \mathrm{H}^{+}$$ $$K_a=\frac{[\mathrm{A}^{-}] [\mathrm{H}^{+}]}{[\mathrm{HA}]}$$ For the aniline reaction, we have Kb: $$\mathrm{A}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HA} + \mathrm{OH}^{-}$$ $$K_b=\frac{[\mathrm{HA}] [\mathrm{OH}^{-}]}{[\mathrm{A}^{-}]}$$ For the given equilibrium: $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}$$ c) We can find the equilibrium constant for the given reaction by dividing the product of \(K_{b}\) of aniline and \(K_w\) by the \(K_{a}\) of benzoic acid: \[K = \frac{K_{b} \times K_{w}}{K_{a}}\] Calculate the value of the equilibrium constant K for the given equilibrium.

Key Concepts

Conjugate Acid-Base Pairs

In acid-base chemistry, substances that differ by a single proton (H⁺) are referred to as conjugate acid-base pairs. Consider benzoic acid \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\). When it donates a proton, it becomes its conjugate base, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\). Similarly, for aniline \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), when it accepts a proton, it forms its conjugate acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\). These pairs illustrate the reversible nature of proton transfer in chemical reactions.
The strength of acids and bases partly depends on the stability of these conjugate partners. A strong acid will have a weak conjugate base, while a strong base will have a weak conjugate acid. This balance is crucial to understand chemical equilibria and reactivity.

Equilibrium Constant

Equilibrium constants give us insights into the position of equilibrium in a chemical reaction. They are denoted by \(K\) and describe the ratio of the concentrations of products to reactants at equilibrium. In this exercise, we examine the equilibrium involving benzoic acid and aniline:

• \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)
• The equilibrium constant \(K\) for this reaction can be calculated using the formula: \[ K = \frac{K_{b} \times K_{w}}{K_{a}} \]
• Where \(K_{w}\) is the ion product of water, \(K_{a}\) and \(K_{b}\) are the acid and base dissociation constants, respectively.

The value of \(K\) helps predict whether the reactants or products are favored at equilibrium, and is a fundamental concept in predicting reaction direction.

pH Calculation

pH is a measure of acidity or basicity of a solution. It is calculated using the formula: \( \mathrm{pH} = -\log[\mathrm{H}^{+}] \). In the exercise, we compare the acidity of benzoic acid and anilinium chloride solutions.

Benzoic Acid Solution

Given \(K_{a} = 6.3 \times 10^{-5}\), the formula to calculate pH is: \[ \mathrm{pH} = -\log(\sqrt{0.10 \times 6.3 \times 10^{-5}}) \]

Anilinium Chloride Solution

For anilinium chloride, since it dissociates into anilinium ions, use:

• \( K_{b} = \frac{1 \times 10^{-14}}{4.3 \times 10^{-10}} = 2.3 \times 10^{-5} \)

Calculate \( \mathrm{pOH} = -\log(\sqrt{0.10 \times 2.3 \times 10^{-5}}) \), then find pH by \( \mathrm{pH} = 14 - \mathrm{pOH} \).
Comparing the pH values, the lower value indicates greater acidity, helping you determine which solution is more acidic.

Acid and Base Strength

The strength of an acid or base is determined by its ability to donate or accept protons. Strong acids completely dissociate in water, while weak acids only partially dissociate.
For benzoic acid, with \(K_{a} = 6.3 \times 10^{-5}\), it's a weak acid. It does not fully dissociate, making its conjugate base relatively stronger compared to the conjugate base of a strong acid.
Aniline, with a very low \(K_{a} = 4.3 \times 10^{-10}\), is also weak but acts as a base. Strong bases like NaOH have high \(K_{b}\) values, unlike aniline.
Understanding these strengths aids in predicting reaction behaviors and equilibrium in acid-base chemistry. Key takeaways include how weaker acids form more stable conjugate bases, and vice versa.