Textbook Question
Identify the stronger base in each pair. Explain your choice. Citing pKa values is not an acceptable answer.
(d)
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Ch. 4 - Acids and Bases: Electron Flow
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
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Table of contents
- Ch. 2 - General Chemistry Translated: Finding the Electrons114
- Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis109
- Ch. 4 - Acids and Bases: Electron Flow144
- Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics118
- Ch. 6 - Stereoisomerism: Arrangement of Atoms in Space112
- Ch. 7 - Principles of Green (Organic) Chemistry3
- Ch. 8 - Alkenes I: Properties and Electrophilic Additions158
- Ch. 9 - Alkenes II: Oxidation and Reduction150
- Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions101
- Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions108
- Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes172
- Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination239
- Ch. 14 - Structural Identification I: Infrared Spectroscopy and Mass Spectrometry54
- Ch. 15 - Structural Identification II: Nuclear Magnetic Resonance Spectroscopy93
- Ch. 16 - Metals in Organic Chemistry48
- Ch. 17 - Carbonyl Addition Reactions: Aldehydes and Ketones45
- Ch. 18 - Nucleophilic Acyl Substitution I: Carboxylic Acids18
- Ch. 19 - Nucleophilic Acyl Substitution II: Carboxylic Acid Derivatives39
- Ch. 20 - Enolates: Carbonyl Addition and Substitution13
- Ch. 21 - Conjugated Systems I: Stability and Addition Reactions56
- Ch. 22 - Conjugated Systems II: Pericyclic Reactions54
- Ch. 23 - Benzene I: Aromatic Stability and Substitution Reactions64
- Ch. 24 - Benzene II: Reactions Influenced by the Aromatic Ring62
- Ch. 25 - Amines: Structure, Reactions, and Synthesis27
- Ch. 26 - Amino Acids, Proteins, and Peptide Synthesis9
- Ch. 27 - Carbohydrates, Nucleic Acids, and Lipids54
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook
Chapter 3, Problem 62h(iii, iv)
For the following acid–base pairs, (iii) predict the favored side of equilibrium; (iv) calculate Keq;
(h) 
Verified step by step guidance1
Step 1: Identify the acid and base on both sides of the equilibrium. In this reaction, NH₃ acts as a base and HCl acts as an acid. The products formed are NH₄⁺ (conjugate acid of NH₃) and Cl⁻ (conjugate base of HCl).
Step 2: Compare the relative strengths of the acids and bases involved. HCl is a strong acid, while NH₄⁺ is a weak acid. Similarly, NH₃ is a weak base, and Cl⁻ is a very weak base.
Step 3: Predict the favored side of the equilibrium. Since the reaction involves a strong acid (HCl) reacting with a weak base (NH₃), the equilibrium will favor the formation of the weaker acid and base, which are NH₄⁺ and Cl⁻.
Step 4: Use the pKa values to calculate the equilibrium constant (Keq). The relationship between Keq and pKa is given by the formula: . Look up the pKa values for HCl and NH₄⁺ to perform this calculation.
Step 5: Interpret the result. A large Keq value indicates that the equilibrium strongly favors the products (NH₄⁺ and Cl⁻), while a small Keq value would indicate that the reactants (NH₃ and HCl) are favored.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acid-Base Equilibrium
Acid-base equilibrium refers to the state in which the rates of the forward and reverse reactions of an acid-base reaction are equal, resulting in constant concentrations of reactants and products. The position of equilibrium can be predicted based on the strength of the acids and bases involved, with stronger acids favoring the formation of weaker acids and vice versa.
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Determining Acid/Base Equilibrium
Equilibrium Constant (K_eq)
The equilibrium constant (K_eq) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. A large K_eq value, such as 1.0 x 10^47, indicates that the products are favored at equilibrium, suggesting that the reaction proceeds almost to completion, while a small K_eq value would indicate that reactants are favored.
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02:19The relationship between equilibrium constant and pKa.
Strength of Acids and Bases
The strength of acids and bases is determined by their ability to donate or accept protons (H+ ions). Strong acids, like HBr, completely dissociate in solution, while weak acids do not. Similarly, strong bases readily accept protons, while weak bases do so less effectively. Understanding the relative strengths of the acids and bases involved is crucial for predicting the favored side of the equilibrium.
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Related Practice
Textbook Question
For the following acid–base pairs, (v) show a mechanism for the reaction; and (vi) draw a reaction coordinate diagram.
(f)
1099
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Textbook Question
Identify the stronger base in each pair. Explain your choice. Citing pKa values is not an acceptable answer.
(a)
1161
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Textbook Question
For the following acid–base pairs, (i) complete the reaction; (ii) identify the acid (A), base (B), conjugate acid (CA), and conjugate base (CB);
(f)
1181
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Textbook Question
Identify the stronger base in each pair. Explain your choice. Citing pKa values is not an acceptable answer.
(c)
1114
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Textbook Question
For the following acid–base pairs, (iii) predict the favored side of equilibrium; (iv) calculate ;
(f)
866
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