Question

In glycine, the carboxylic acid group has \(K_{\mathrm{a}}=4.3 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=6.0 \times 10^{-5}\). Use these equilibrium constant values to calculate the equilibrium constants for the following. a. \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) b. \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{OH}^{-}\) c. \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}\)

Short answer

The equilibrium constants for the three reactions are: a. \(K_{1} = 6.0 \times 10^{-5}\) b. \(K_{2} = 4.3 \times 10^{-3}\) c. \(K_{3} = 2.58 \times 10^{-7}\)

Step-by-step solution

Determine the relationship between Ka, Kb, and Kw

From the relationship of Ka and Kb, we know that the product of Ka and Kb equals Kw (the ion-product constant for water, which is approximately \(1.0 \times 10^{-14}\)): \[K_{a} \times K_{b} = K_{w}\] For the reactions given, we will be using this relationship in order to find the equilibrium constants.

Calculate the equilibrium constant for reaction (a)

Reaction (a): \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) In this reaction, the amino group loses a proton and forms water as a base. Therefore, we need to use \(K_b\) value for the amino group: \[K_{b} = 6.0 \times 10^{-5}\] The expression for the equilibrium constant of reaction (a) can be written as \(K_{1} = \frac{[H_{2}NCH_{2}CO_{2}^{-}][H_{3}O^{+}]}{[^{+}H_{3}NCH_{2}CO_{2}^{-}][H_{2}O]}\). Since the concentration of water is constant, the equilibrium constant for reaction (a) will simply be equal to the given Kb value: \[K_{1} = K_{b} = 6.0 \times 10^{-5}\]

Calculate the equilibrium constant for reaction (b)

Reaction (b): \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{OH}^{-}\) In this reaction, the carboxylic acid group acts as an acid and donates a proton. Therefore, we need to use \(K_a\) value for the carboxylic acid group: \[K_{a} = 4.3 \times 10^{-3}\] The expression for the equilibrium constant of reaction (b) can be written as \(K_{2} = \frac{[H_{2}NCH_{2}CO_{2}H][OH^{-}]}{[H_{2}NCH_{2}CO_{2}^{-}][H_{2}O]}\). Again, since the concentration of water is constant, the equilibrium constant for reaction (b) will be equal to the given Ka value: \[K_{2} = K_{a} = 4.3 \times 10^{-3}\]

Calculate the equilibrium constant for reaction (c)

Reaction (c): \({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}\) This reaction is the combined form of reactions (a) and (b). Also, it is a combination of acid and base equilibria. We need to find the equilibrium constant for this reaction, which we will call \(K_{3}\). To do this, we can multiply the equilibrium constants of reactions (a) and (b): \[K_{3} = K_{1} \times K_{2} = (6.0 \times 10^{-5}) \times (4.3 \times 10^{-3})\] \[K_{3} = 2.58 \times 10^{-7}\] The equilibrium constants for the three reactions are: a. \(K_{1} = 6.0 \times 10^{-5}\) b. \(K_{2} = 4.3 \times 10^{-3}\) c. \(K_{3} = 2.58 \times 10^{-7}\)

Key concepts

Carboxylic Acid Group

When it comes to the chemistry of organic molecules, the carboxylic acid group plays a significant role. Structurally, it's represented as \( -COOH \), featuring both a carbonyl group (\(C=O\)) and a hydroxyl group (\(OH\)). This functional group is known for being acidic due to its ability to donate a proton (\(H^+\)) to a base, which makes it fundamentally important in acid-base chemistry.

Regarding the exercise, the carboxylic acid group of glycine exhibits a specific equilibrium constant for the dissociation \( (K_{a}) \). The \( K_{a} \) value, \(4.3 \times 10^{-3}\), quantifies the strength of the acid in water, essentially telling us how readily the carboxylic acid group gives up its proton to become negatively charged \( (COO^-) \). This property is crucial when predicting the behavior of glycine in various pH environments and plays an integral part in amino acid chemistry.

Amino Group

The amino group, typically represented as \( -NH_2 \), is another fundamental functional group in organic chemistry, especially in the structure of amino acids like glycine. It consists of a nitrogen atom attached to hydrogen atoms and can act as a base by accepting a proton (\( H^+ \)) to form an ammonium ion (\( NH_3^+ \)).

In the reaction presented, the amino group of glycine possesses its own equilibrium constant known as the base dissociation constant \( (K_{b}) \), which is \(6.0 \times 10^{-5}\). This value measures the avidity of the amino group to gain a proton from water. Understanding the behavior of the amino group in reactions is key to many biological and chemical systems, including the formation of proteins from amino acids.

Ion-Product Constant of Water

Diving into the importance of the ion-product constant of water \( (K_{w}) \), we find it represents the equilibrium constant for the self-ionization of water, which is \( H_2O \leftrightarrow H^+ + OH^- \). The value of \(K_{w}\) is approximately \(1.0 \times 10^{-14}\) at standard temperature, outlining the relationship between the hydrogen ion concentration \([H^+]\) and the hydroxide ion concentration \([OH^-]\) in pure water.

This constant is instrumental in acid-base equilibrium as it is a fixed value under given conditions and allows chemists to connect the acid dissociation constants (\(K_{a}\)) and base dissociation constants (\(K_{b}\)) of conjugate acid-base pairs. In our example of glycine, it is used to understand how the carboxylic acid and amino groups behave in water, informing us about their concurrent acid-base equilibria.

Equilibrium Constant Calculation

The equilibrium constant calculation is a quantitative expression of the ratio of concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. It plays a pivotal role in chemical reactions as it allows chemists to predict the extent of a reaction and the concentrations of different species at equilibrium.

In the context of our glycine scenario, we see how the carboxylic acid and amino groups have defined equilibrium constants associated with their acid and base dissociation. By understanding the relationships between these values and the ion-product constant of water, we can solve for the equilibrium constants of individual reactions, providing insights into the prevalent forms of glycine under various conditions. Specifically, the multiplication of \(K_{a}\) and \(K_{b}\) to solve for reaction (c) is a reflection of the reaction being a sum of two equilibria, neatly illustrating the interconnectedness of equilibrium concepts in chemistry.