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Ch.16 - Acid-Base Equilibria
Chapter 16, Problem 114a

The amino acid glycine (H2N–CH2–COOH) can participate in the following equilibria in water:
H2N–CH2–COOH + H2O ⇌ H2N–CH2–COO + H3O+ Ka = 4.3 × 10-3
H2N–CH2–COOH + H2O⇌ +H3N–CH2–COOH + OH- Kb = 6.0 × 10-5
(a) Use the values of Ka and Kb to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: H2N–CH2–COOH ⇌ +H3N–CH2–COO

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Identify the relevant equilibria and their constants. For glycine, the acid dissociation constant (Ka) for the reaction H2N-CH2-COOH + H2O ⇌ H2N-CH2-COO- + H3O+ is given as 4.3 * 10^-3. The base dissociation constant (Kb) for the reaction H2N-CH2-COOH + H2O ⇌ H3N-CH2-COOH + OH- is given as 6.0 * 10^-5.
Understand the relationship between Ka, Kb, and Kw (the ion-product constant of water, which is typically 1.0 * 10^-14 at 25°C). The relationship is given by the equation Ka * Kb = Kw.
Calculate the equilibrium constant (K) for the formation of the zwitterion, H2N-CH2-COOH ⇌ H3N-CH2-COO-, by using the relationship K = Kw / (Ka * Kb). This equation arises because the formation of the zwitterion can be thought of as a combination of the two given reactions, where both the acid and base properties of glycine are utilized.
Substitute the given values of Ka and Kb into the equation to estimate K. Use Kw = 1.0 * 10^-14, Ka = 4.3 * 10^-3, and Kb = 6.0 * 10^-5.
Interpret the result. A larger value of K indicates a greater extent of zwitterion formation at equilibrium, suggesting that the intramolecular proton transfer is favorable under the given conditions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acid-Base Equilibria

Acid-base equilibria involve the transfer of protons (H+) between species in solution. In this context, glycine can act as both an acid and a base, donating or accepting protons. The equilibrium constants, Ka and Kb, quantify the strength of these acid-base reactions, indicating how readily the species dissociate or associate in water.
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Arrhenius Acids and Bases

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction. For acid-base reactions, K can be derived from the dissociation constants (Ka and Kb) of the species involved. Understanding how to manipulate these constants is essential for calculating the equilibrium position of related reactions, such as the formation of zwitterions.
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Zwitterions

Zwitterions are molecules that contain both positive and negative charges but are overall electrically neutral. In the case of glycine, the zwitterionic form is formed when the amino group (NH2) accepts a proton, while the carboxyl group (COOH) donates a proton. This unique structure plays a crucial role in the behavior of amino acids in solution, particularly in their acid-base equilibria.
Related Practice
Textbook Question

The following observations are made about a diprotic acid H2A: (i) A 0.10 M solution of H2A has pH = 3.30. (ii) A 0.10 M solution of the salt NaHA is acidic. Which of the following could be the value of pKa2 for H2A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82?

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Open Question
Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that the pH in the stomach is 2.5, indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, Kb = 7 * 10-7; caffeine, Kb = 4 * 10-14; strychnine, Kb = 1 * 10-6; quinine, Kb = 1.1 * 10-6.
Open Question
The amino acid glycine H₂N¬CH₂¬COOH can participate in the following equilibria in water: H₂N¬CH₂¬COOH + H₂O ⇌ H₂N¬CH₂¬COO⁻ + H₃O⁺ with Ka = 4.3 × 10⁻³, and H₂N¬CH₂¬COOH + H₂O ⇌ +H₃N¬CH₂¬COOH + OH⁻ with Kb = 6.0 × 10⁻⁵. (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1?
Textbook Question

The amino acid glycine (H2N–CH2–COOH) can participate in the following equilibria in water:

H2N–CH2–COOH + H2O ⇌ H2N–CH2–COO + H3O+ Ka = 4.3 × 10-3

H2N–CH2–COOH + H2O⇌ +H3N–CH2–COOH + OH- Kb = 6.0 × 10-5

(b) What is the pH of a 0.050 M aqueous solution of glycine?

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Open Question
The pKb of water is _______. (a) 1 (b) 7 (c) 14 (d) not defined (e) none of the above
Open Question
Calculate the number of H3O+ ions in 1.0 mL of pure water at 25 °C.