Question

Predict the position of equilibrium and calculate the equilibrium constant, \(K_{\text {eq }}\), for each acid-base reaction. (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) \(\begin{array}{ccc}\text { Methylamine } \quad \text { Acetic acid } & \text { Methylammonium } & \text { Acetate } \\ \text { ion } & \text { ion }\end{array}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NH}_{2}^{-}\) Ethoxide ion Ammonia Ethanol Amide ion

Short answer

In summary, the first reaction between methylamine and acetic acid has an equilibrium constant of \(K_\text{eq}(\text{a}) = 1.26 \times 10^{-6}\) and the equilibrium lies to the right, favoring the products. The second reaction between the ethoxide ion and ammonia has an equilibrium constant of \(K_\text{eq}(\text{b}) = 1.79\) and the equilibrium lies to the left, favoring the reactants.

Step-by-step solution

Identify the acids and bases in each reaction

For the first reaction, we can see that methylamine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)) is the base and acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) is the acid. In the second reaction, the ethoxide ion (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}^{-}\)) is the base and ammonia (\(\mathrm{NH}_{3}\)) is the acid.

Determine the position of equilibrium

We can determine the position of equilibrium by comparing the strengths of the acids and bases involved in each reaction. In general, the equilibrium will lie to the side of the weaker acid and weaker base. In the first reaction, acetic acid is a weaker acid than methylammonium ion (\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)), and methylamine is a stronger base than acetate ion (\(\mathrm{CH}_{3}\mathrm{COO}^{-}\)), so the equilibrium will lie to the right, favoring the formation of products. In the second reaction, ammonia is a weaker acid than ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)), and the ethoxide ion is a stronger base than the amide ion (\(\mathrm{NH}_{2}^{-}\)), so the equilibrium will lie to the left, favoring the reactants.

Obtain the \(K_{\text {a }}\) values for the acids

We will use the following \(K_{\text {a }}\) values for acetic acid and ammonia, which can be found in a table or other reference sources: \(K_{\text {a }}\) for acetic acid = \(1.8 \times 10^{-5}\) \(K_{\text {a }}\) for ammonia = \(5.6 \times 10^{-10}\)

Calculate the \(K_{\text {eq }}\) values for each reaction

To calculate the equilibrium constant \(K_{\text {eq }}\) for each reaction, use the formula: \(K_\text{eq} = \frac{K_\text{a}(\text{conjugate acid})}{K_\text{a}(\text{original acid})}\) For the first reaction: \(K_\text{eq}(\text{a}) = \frac{K_\text{a}(\text{methylammonium ion})}{K_\text{a}(\text{acetic acid})}\) \(K_\text{a}\) for methylammonium ion can be calculated using methylamine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)) as a base and considering \(K_\text{b} = 4.4 \times 10^{-4}\): \(K_\text{a}(\text{methylammonium ion}) = \frac{K_\text{w}}{K_\text{b}(\text{methylamine})} = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}} = 2.27 \times 10^{-11}\) \(K_\text{eq}(\text{a}) = \frac{2.27 \times 10^{-11}}{1.8 \times 10^{-5}} = 1.26 \times 10^{-6}\) For the second reaction: \(K_\text{eq}(\text{b}) = \frac{K_\text{a}(\text{ethanol})}{K_\text{a}(\text{ammonia})}\) \(K_\text{a}\) for ethanol can be calculated using ethoxide ion (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}^{-}\)) as the base and considering \(K_\text{b} = 1 \times 10^{-5}\): \(K_\text{a}(\text{ethanol}) = \frac{K_\text{w}}{K_\text{b}(\text{ethoxide ion})} = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9}\) \(K_\text{eq}(\text{b}) = \frac{1 \times 10^{-9}}{5.6 \times 10^{-10}} = 1.79\) Thus, the equilibrium constants for the two reactions are \(K_\text{eq}(\text{a}) = 1.26 \times 10^{-6}\) and \(K_\text{eq}(\text{b}) = 1.79\).

Key concepts

Equilibrium Constant Calculation

Understanding the equilibrium constant calculation is a fundamental part of mastering acid-base reactions. The equilibrium constant, denoted as K_eq, is a numerical value that indicates the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their respective coefficients as represented in the balanced chemical equation.

In our example, to calculate K_eq for an acid-base reaction, we use the relationship between the K_a values of the acids involved. For a generic reaction where acid HA becomes its conjugate base A^-, and base B becomes its conjugate acid BH⁺, the K_eq expression would be:K_eq = [BH⁺][A^-] / [HA][B].However, if we know the K_a values of the acids (and K_b for the bases, when necessary), we can simplify by using K_eq = K_a(conjugate acid) / K_a(original acid) as shown in the step-by-step solution.

Acid-Base Reaction Prediction

Predicting the direction in which an acid-base reaction will proceed is intrinsic to understanding chemical equilibrium. The principle guiding this prediction is that a reaction will favor the formation of the weaker acid and the weaker base. So, when comparing two acids, we consider their K_a values: the one with the lower K_a is the weaker acid and will be favored in the product side of the equilibrium equation.

In the given examples, we analyze the strengths of the acids and bases in each reaction set. A stronger base such as methylamine will tend to accept a proton more readily than the weaker acetate ion, pushing the reaction towards the right. Conversely, since ammonia is a weaker acid than ethanol, the position of equilibrium in the second reaction leans towards the reactants' side.

Conjugate Acids and Bases

Conjugate acids and bases are a pair of substances that transform into each other by the gain or loss of a proton (H⁺). Every acid has its conjugate base, formed when it donates a proton, and every base has its conjugate acid, formed when it accepts a proton.

Understanding these pairs is crucial for various chemical calculations and predictions. For instance, in the reaction involving methylamine, we see that when it gains a proton, it becomes the conjugate acid methylammonium ion. Similarly, acetic acid, upon losing a proton, becomes the conjugate base acetate ion. The strength of a conjugate acid-base pair is inversely related, meaning if we have a strong acid, its conjugate base is weak, and vice versa.

Ka and Kb Values

The K_a value, or the acid dissociation constant, quantifies the strength of an acid in solution; it refers to the acid's ability to donate protons. The K_b value, or the base dissociation constant, similarly quantifies the strength of a base to accept protons. These values are particularly important when calculating equilibrium constants for acid-base reactions.

Leaning on these values, chemists can foresee the equilibrium position of an acid-base reaction. If we know the K_b of a base (like methylamine with its K_b of 4.4 x 10^-4), we can calculate its conjugate acid's K_a by using the ion product constant for water (K_w) with the formula K_a = K_w / K_b. These computed K_a values of the conjugate acids can then be used, along with the acid K_a values, to find the K_eq of the overall reaction.