Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

3. Acids and Bases

Equilibrium Constant

Organic Chemistry

3. Acids and Bases

Equilibrium Constant

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Problem 63b

Textbook Question

Calculate K_eq for the following acid–base reaction.

Verified step by step guidance

Step 1: Identify the acid and base on both sides of the reaction. In this reaction, the methyl anion (CH₃⁻) acts as the base, and isopropanol (CH₃CHOHCH₃) acts as the acid on the left side. On the right side, methane (CH₄) is the conjugate acid, and the isopropoxide ion (CH₃CH⁻OCH₃) is the conjugate base.

Step 2: Recall the relationship between pKa values and equilibrium constants (Keq). The equilibrium constant for an acid-base reaction can be calculated using the formula: Keq = 10^(pKa(acid on left) - pKa(acid on right)).

Step 3: Look up the pKa values for the acids involved. The pKa of isopropanol (CH₃CHOHCH₃) is approximately 16. The pKa of methane (CH₄) is approximately 50.

Step 4: Substitute the pKa values into the formula for Keq. Using the formula Keq = 10^(pKa(acid on left) - pKa(acid on right)), substitute pKa(acid on left) = 16 and pKa(acid on right) = 50.

Step 5: Simplify the expression to determine the magnitude of Keq. The calculation will show that Keq is a very small number, indicating that the equilibrium strongly favors the reactants (left side of the reaction).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons (H+) between reactants. In organic chemistry, these reactions are fundamental for understanding the behavior of various compounds, particularly in determining their reactivity and stability. The strength of acids and bases is often quantified using the pKa scale, which helps predict the direction of the reaction.

Equilibrium Constant (K_eq)

The equilibrium constant (K_eq) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction. It provides insight into the favorability of a reaction; a K_eq greater than 1 indicates that products are favored, while a K_eq less than 1 suggests that reactants are favored. Calculating K_eq is essential for understanding the extent of acid-base reactions.

Relationship Between pKa and K_eq

The relationship between pKa and K_eq is crucial in acid-base chemistry. The pKa of an acid is the negative logarithm of its acid dissociation constant (Ka), and it can be used to derive K_eq for acid-base reactions. Specifically, K_eq can be calculated using the difference in pKa values of the acids and bases involved, allowing for a quantitative assessment of the reaction's equilibrium position.

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Related Practice

Textbook Question

For each of the following compounds (here shown in their acidic forms), write the form that predominates in a solution with a pH = 5.5:

a. CH₃COOH(pK_a= 4.76)

b. CH₃CH₂N⁺H₃ (pK_a= 11.0)

Textbook Question

At what pH is the concentration of a compound, with a pK_a = 8.4, 100 times greater in its acidic form than in its basic form? At what pH is 50% of a compound, with a pK_a = 7.3, in its basic form?

Textbook Question

If an acid with a pK_a of 5.3 is in an aqueous solution of pH 5.7, what percentage of the acid is present in its acidic form?

Textbook Question

For each of the following compounds (here shown in their acidic forms), write the form that predominates in a solution with a pH = 5.5:

i. HON⁺H₃ (pK_a = 6.0)

Textbook Question

For each of the following compounds (here shown in their acidic forms), write the form that predominates in a solution with a pH = 5.5:

g. HNO₂ (pK_a= 3.4)

h. HNO₃ (pK_a= −1.3)

Textbook Question

As long as the pH is not less than ___________, at least 50% of a protonated amine with a pK_a value of 10.4 will be in its neutral, nonprotonated form.

Textbook Question

Indicate whether a protonated amine (RN⁺H₃) with a pKa value of 9 has more charged or more neutral molecules in a solution with the pH values given in Problem 41.

1. pH = 1

2. pH = 3

3. pH = 5

4. pH = 7

5. pH = 10

6. pH = 13

Textbook Question

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 (pK_a= 4.9)

2. CH₃N⁺H₃ (pK_a= 10.7)

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