Textbook Question
Given the following Keq values, how much stronger of a base is acetate (CH3CO2-) than chloride (Cl⁻)?
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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 30b
Given that the indicated pKa values correspond to the acid dissociation reactions shown, calculate the ratio of acid to conjugate base for the reactions shown.
(b) 
Verified step by step guidance1
Step 1: Understand the relationship between pKₐ and the acid dissociation constant (Kₐ). The pKₐ value is defined as pKₐ = -log(Kₐ). This value helps determine the strength of the acid and its dissociation behavior in solution.
Step 2: Recall the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKₐ and the ratio of acid to conjugate base: \( \text{pH} = \text{pKₐ} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \([\text{HA}]\) represents the concentration of the acid, and \([\text{A}^-]\) represents the concentration of the conjugate base.
Step 3: Rearrange the Henderson-Hasselbalch equation to solve for the ratio of acid to conjugate base: \( \frac{[\text{HA}]}{[\text{A}^-]} = 10^{\text{pKₐ} - \text{pH}} \). This equation allows you to calculate the ratio directly once the pH and pKₐ values are known.
Step 4: Substitute the given pKₐ value and the pH of the solution into the rearranged equation. Ensure that the pH value corresponds to the conditions of the reaction being analyzed.
Step 5: Perform the calculation using the substituted values to determine the ratio of acid to conjugate base. The result will indicate the relative concentrations of the acid and its conjugate base in the solution.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
pKₐ and Acid-Base Equilibrium
pKₐ is a measure of the strength of an acid in solution, defined as the negative logarithm of the acid dissociation constant (Kₐ). It indicates how readily an acid donates protons (H⁺) to form its conjugate base. A lower pKₐ value signifies a stronger acid, which dissociates more completely in solution. Understanding pKₐ is essential for calculating the ratio of acid to conjugate base in equilibrium.
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Determining Acid/Base Equilibrium
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a solution to the pKₐ of an acid and the ratio of the concentrations of its conjugate base and acid. It is expressed as pH = pKₐ + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the acid. This equation is crucial for determining the desired ratio of acid to conjugate base in buffer solutions.
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Acid-Base Reaction Dynamics
Acid-base reactions involve the transfer of protons between species, leading to the formation of conjugate acids and bases. The position of equilibrium in these reactions is influenced by the relative strengths of the acids and bases involved, as well as their concentrations. Understanding these dynamics is vital for calculating the ratios of acid to conjugate base, as it helps predict how changes in concentration or pH will affect the equilibrium state.
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Related Practice
Textbook Question
Given the following Keq values, how much stronger of an acid is methanol (CH3OH) than acetylene (HC≡CH)?
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Textbook Question
Hydrogen gas (H2) has a relatively high pKa value. Is it a stable or unstable acid? Do you expect it to participate in acid–base reactions?
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Textbook Question
If a base has a conjugate acid with a high pKa value, is it stable or unstable? How do you know this is true (besides 'I just remember')?
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Textbook Question
Given that the indicated pKa values correspond to the acid dissociation reactions shown, calculate the ratio of acid to conjugate base for the reactions shown.
(a)
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Textbook Question
How do you know that a proton with a low pKa value is acidic (besides 'I just remember')?
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