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.
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 72a
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 72aChapter 3, Problem 72a
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.
Verified step by step guidance1
Identify the structure of citric acid from the image. Citric acid contains three carboxylic acid (COOH) groups, with pKa values labeled as 3.1, 4.5, and 5.8. The central COOH group has the lowest pKa (3.1), indicating it is the most acidic.
Understand the concept of pKa. The pKa value is inversely related to the acidity of a compound; a lower pKa means the compound is more acidic. Acetic acid, for comparison, has a pKa of approximately 4.76.
Analyze the electronic environment of the central COOH group. The central COOH group is surrounded by two other carboxylic acid groups. These groups are electron-withdrawing due to their electronegative oxygen atoms, which stabilize the negative charge on the conjugate base formed after deprotonation.
Explain the inductive effect. The electron-withdrawing nature of the adjacent COOH groups increases the acidity of the central COOH group by stabilizing its conjugate base. This effect is stronger for the central COOH group compared to acetic acid, which has no such electron-withdrawing groups nearby.
Conclude that the lower pKa of the central COOH group in citric acid compared to acetic acid is due to the inductive effect of the neighboring carboxylic acid groups, which enhance the acidity of the central COOH group.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acidity and pKa
The pKa value is a measure of the acidity of a compound, indicating the strength of an acid in solution. A lower pKa value signifies a stronger acid, meaning it dissociates more readily to release protons (H+). In the context of citric acid, the presence of multiple COOH groups can influence its overall acidity compared to simpler acids like acetic acid.
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Identifying pKa values
Structural Effects on Acidity
The structure of a molecule significantly affects its acidity. In citric acid, the arrangement of the three carboxylic acid (COOH) groups can create intramolecular interactions that stabilize the negative charge on the conjugate base after deprotonation. This stabilization can lead to a lower pKa for the central COOH group compared to acetic acid, which has only one COOH group.
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Comparative Acidity of Carboxylic Acids
When comparing carboxylic acids, factors such as electronegativity, resonance, and inductive effects play crucial roles in determining acidity. Citric acid, with its multiple COOH groups, exhibits stronger electron-withdrawing effects and resonance stabilization in its conjugate base, making it more acidic than acetic acid, which lacks these additional stabilizing features.
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Related Practice
Textbook Question
Given that pH + pOH = 14 and that the concentration of water in a solution of water is 55.5 M, show that the pKa of water is 15.7. (Hint: pOH=−log [HO−])
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Textbook Question
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)
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Textbook Question
Which of the four reactions has the most favorable equilibrium constant?
1.
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Textbook Question
How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH.
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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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