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?

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 \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\).

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.

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.