Textbook Question
From which of the following compounds can HO− remove a proton in a reaction that favors product formation?
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Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
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Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 69
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 69Chapter 3, Problem 69
You are planning to carry out a reaction that produces protons. The reaction will be buffered at pH = 10.5. Would it be better to use a protonated methylamine/methylamine buffer or a protonated ethylamine/ethylamine buffer? (pKa of protonated methylamine = 10.7; pKa of protonated ethylamine = 11.0)
Verified step by step guidance1
Step 1: Understand the concept of buffering capacity. A buffer works best when the pH of the solution is close to the pKa of the conjugate acid in the buffer system. This is because the buffer can effectively resist changes in pH when the concentrations of the weak acid and its conjugate base are similar.
Step 2: Compare the pKa values of the two buffer systems provided. The pKa of protonated methylamine is 10.7, and the pKa of protonated ethylamine is 11.0. The closer the pKa is to the desired pH (10.5), the better the buffer system will perform.
Step 3: Calculate the difference between the pH and the pKa for each buffer system. For the protonated methylamine/methylamine buffer, the difference is |10.5 - 10.7| = 0.2. For the protonated ethylamine/ethylamine buffer, the difference is |10.5 - 11.0| = 0.5.
Step 4: Based on the calculations, the protonated methylamine/methylamine buffer has a smaller difference between its pKa and the desired pH, meaning it is closer to the optimal buffering range.
Step 5: Conclude that the protonated methylamine/methylamine buffer would be a better choice for buffering at pH 10.5, as it is more effective at maintaining the desired pH due to its pKa being closer to the target pH.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Buffer Solutions
Buffer solutions are mixtures that resist changes in pH upon the addition of small amounts of acids or bases. They typically consist of a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, the choice between methylamine and ethylamine buffers is crucial for maintaining the desired pH during the reaction.
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pKa and Acid-Base Equilibrium
The pKa value is a measure of the strength of an acid in solution; it indicates the pH at which half of the acid is dissociated. A lower pKa value means a stronger acid. In this scenario, the pKa values of the protonated forms of methylamine and ethylamine help determine which buffer will be more effective at pH 10.5, as the buffer should ideally have a pKa close to the target pH.
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Protonation and Deprotonation
Protonation refers to the addition of a proton (H+) to a molecule, while deprotonation is the removal of a proton. The balance between these processes is essential in buffer systems, as they determine the concentration of protonated and deprotonated species. In this question, understanding how methylamine and ethylamine behave at the specified pH will influence the choice of buffer for effective proton management.
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Related Practice
Textbook Question
For each of the following pairs of reactions, indicate which one has the more favorable equilibrium constant (that is, which one most favors products):
1.
2.
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Textbook Question
a. Without using a calculator, estimate the pH of each of the following solutions:
1. [HO−] = 3.2 × 10−5
2. [H3O+] = 8.3 × 10−1
3. [H3O+] = 1.7 × 10−3
b. Determine the exact pH, using a calculator.
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Textbook Question
Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following:
a. The first pKa (for the COOH group in the center of the molecule) is lower than the pKa of acetic acid.
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Textbook Question
Which of the four reactions has the most favorable equilibrium constant?
1.
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Textbook Question
Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following:
b. The third pKa is greater than the pKa of acetic acid.
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