Question

Predict the position of equilibrium and calculate the equilibrium constant, \(K_{\mathrm{eq}}\) ' for each acid-base reaction. (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NH}_{2}^{-}\)

Short answer

Question: Predict the position of equilibrium and calculate the equilibrium constant (\(K_{eq}\)) for the following acid-base reactions: (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2} + \mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} + \mathrm{CH}_{3} \mathrm{COO}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{NH}_{2}^{-}\) Answer: The position of equilibrium for both reactions lies towards the right side, favoring the formation of the conjugate acid and conjugate base. The equilibrium constants are \(K_{eq} = 6.87\) for the first reaction and \(K_{eq} = 27.8\) for the second reaction.

Step-by-step solution

Identify the acids and bases in each reaction

In each given reaction, identify the acid and the base on the left-hand side. The acid will be the species that donates the proton (H+), and the base will be the species that accepts the proton. The products of the reaction will be the conjugate acid (formed after the base accepts the proton) and the conjugate base (formed after the acid donates the proton). (a) In the first reaction: \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is the base, and \(\mathrm{CH}_{3} \mathrm{COOH}\) is the acid. (b) In the second reaction: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is the base, and \(\mathrm{NH}_{3}\) is the acid.

Determine the position of equilibrium

Compare the relative strengths of the acids and bases involved in the reactions. Generally, a stronger acid will react with a stronger base to form a weaker acid and a weaker base. The position of equilibrium will lie towards the side with the weaker acid and weaker base. (a) In the first reaction: the \(\mathrm{CH}_{3} \mathrm{COOH}\) is a relatively weak acid, and the \(\mathrm{CH}_3 \mathrm{NH}_2\) is a moderately strong base. Therefore, the position of equilibrium is expected to lie towards the right side, favoring the formation of the conjugate acid \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\) and conjugate base \(\mathrm{CH}_{3} \mathrm{COO}^{-}\). (b) In the second reaction: \(\mathrm{NH}_{3}\) is a weaker base compared to \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\), and the weaker acid is the \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Therefore, the position of equilibrium is expected to lie towards the right side, favoring the formation of the conjugate acid \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and the conjugate base \(\mathrm{NH}_{2}^{-}\).

Calculate the equilibrium constant \(K_{eq}\) for each reaction

First, determine the \(K_{a}\) values of the acids, and the \(K_{b}\) values of the bases from a table or other reference source. Then, we can calculate the equilibrium constant using the relationship \(K_{eq} = \frac{K_{a} * K_{b}}{K_{w}}\), where \(K_{w}\) is the ion product of water, which is \(1.0 \times 10^{-14}\) at 25°C. (a) For the first reaction: \(K_{a}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) : \(1.74 \times 10^{-5}\) \(K_{b}\) of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) : \(3.95 \times 10^{-5}\) Calculate \(K_{eq}\) for the reaction: \(K_{eq} = \frac{1.74 \times 10^{-5} * 3.95 \times 10^{-5}}{1.0 \times 10^{-14}} \approx 6.87\) (b) For the second reaction: \(K_{a}\) of \(\mathrm{NH}_{3}\) : \(5.56 \times 10^{-10}\) \(K_{b}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) : \(5.0 \times 10^{-6}\) Calculate \(K_{eq}\) for the reaction: \(K_{eq} = \frac{5.56 \times 10^{-10} * 5.0 \times 10^{-6}}{1.0 \times 10^{-14}} \approx 27.8\) The equilibrium constants are 6.87 for the first reaction and 27.8 for the second reaction, supporting our predictions in Step 2 that the position of equilibrium is towards the right side for both reactions.