Domanda del libro di testo
For each of the following compounds, indicate the pH at which
a. 50% of the compound is in a form that possesses a charge.
1. CH3CH2COOH (pKa = 4.9)
2. CH3N+H3 (pKa = 10.7)
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Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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TOC.tableOfContents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding31
- 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 ChemistryProblema 45a
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblema 45aCapitolo 2, Problema 45a
At what pH is the concentration of a compound, with a pKa = 8.4, 100 times greater in its acidic form than in its basic form? At what pH is 50% of a compound, with a pKa = 7.3, in its basic form?
Guida verificata passo dopo passo1
Step 1: Understand the relationship between pH, pKa, and the ratio of acidic to basic forms of a compound using the Henderson-Hasselbalch equation: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \([\text{A}^-]\) represents the concentration of the basic form, and \([\text{HA}]\) represents the concentration of the acidic form.
Step 2: For the first part of the problem, where the concentration of the acidic form is 100 times greater than the basic form, set up the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} = \frac{1}{100} \). Substitute this ratio into the Henderson-Hasselbalch equation: \( \text{pH} = \text{pKa} + \log \left( \frac{1}{100} \right) \). Simplify \( \log \left( \frac{1}{100} \right) \) to \( -2 \).
Step 3: Substitute the given \( \text{pKa} = 8.4 \) into the equation: \( \text{pH} = 8.4 - 2 \). This will give the pH at which the acidic form is 100 times greater than the basic form.
Step 4: For the second part of the problem, where 50% of the compound is in its basic form, recognize that \( \frac{[\text{A}^-]}{[\text{HA}]} = 1 \). When the concentrations of the acidic and basic forms are equal, \( \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = \log(1) = 0 \).
Step 5: Substitute \( \text{pKa} = 7.3 \) into the Henderson-Hasselbalch equation: \( \text{pH} = \text{pKa} + 0 \). This means the pH at which 50% of the compound is in its basic form is equal to its \( \text{pKa} \).
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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 deprotonated (basic) and protonated (acidic) forms. It is expressed as pH = pKa + log([A-]/[HA]). This equation is essential for determining the pH at which a specific ratio of acidic to basic forms exists, making it crucial for solving the given question.
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pKa and pH Relationship
The pKa value of an acid indicates the pH at which half of the acid is dissociated into its conjugate base. A lower pKa means a stronger acid, while a higher pKa indicates a weaker acid. Understanding this relationship is vital for determining the pH at which a compound exists predominantly in its acidic or basic form, as required in the question.
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Concentration Ratios
In the context of acid-base chemistry, concentration ratios refer to the relative amounts of the acidic and basic forms of a compound. For example, if the acidic form is 100 times greater than the basic form, this translates to a ratio of 100:1. Recognizing how to manipulate these ratios using the Henderson-Hasselbalch equation is key to finding the specific pH values requested in the question.
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Domanda del libro di testo
For each of the following compounds, indicate the pH at which
b. more than 99% of the compound is in a form that possesses a charge.
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