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

Acid Base Equilibrium

Organic Chemistry

3. Acids and Bases

Acid Base Equilibrium

Previous problemNext problem

3:31 minutes

Problem 6a

Textbook Question

An acid has a K_a of 4.53 × 10⁻⁶ in water. What is its K_eq for reaction with water in a dilute solution? ([H₂O] = 55.5 M)

Verified step by step guidance

Step 1: Understand the relationship between Ka and Keq. Ka represents the acid dissociation constant, which is specific to the reaction of the acid with water. Keq, the equilibrium constant, can be calculated using the formula: Keq = Ka × [H2O], where [H2O] is the molar concentration of water in the solution.

Step 2: Identify the given values. From the problem, Ka = 4.53×10⁻⁶ and [H2O] = 55.5 M. These values will be used in the formula to calculate Keq.

Step 3: Write the formula for Keq explicitly using MathML:

K_{eq} = K_{a} \times {[H}_{2O}

Step 4: Substitute the given values into the formula. Replace Ka with 4.53×10⁻⁶ and [H2O] with 55.5 M in the equation.

Step 5: Perform the multiplication to find Keq. Multiply Ka by [H2O] to determine the equilibrium constant for the reaction. Ensure proper handling of significant figures and scientific notation during the calculation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) quantifies the strength of an acid in solution. It is defined as the equilibrium constant for the dissociation of an acid into its conjugate base and a proton. A higher Ka value indicates a stronger acid, as it dissociates more completely in water. In this case, the given Ka of 4.53×10−6 suggests a weak acid that partially ionizes in solution.

Equilibrium Constant (Keq)

The equilibrium constant (Keq) represents the ratio of the concentrations of products to reactants at equilibrium for a given reaction. For the dissociation of an acid in water, Keq can be derived from the Ka value, taking into account the concentration of water, which is typically treated as a constant in dilute solutions. Understanding Keq is essential for predicting the extent of the reaction and the concentrations of species at equilibrium.

Water as a Solvent

Water is often considered a solvent in acid-base reactions due to its high dielectric constant and ability to stabilize ions. In dilute solutions, the concentration of water is approximately 55.5 M, which can be factored into equilibrium calculations. This concentration allows for simplifications in the equilibrium expressions, as the activity of water is effectively constant, influencing the calculation of Keq for the acid's reaction with water.

Watch next

Master The 3 steps for determining the direction of acid and base equilibrium. with a bite sized video explanation from Johnny

Start learning

Related Videos

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:

c. H₃O⁺ (pK_a = −1.7)

d. HBr (pK_a = −9)

Textbook Question

Do the equilibria of the following acid–base reactions lie to the right or the left? (The pK_a of H2O2 is 11.6.)

HOOH + HO⁻⇌ HOO⁻+ H2O

RC☰CH + HOO⁻⇌ RC☰C⁻+ HOOH

Textbook Question

Using pK_a values for the conjugate acids of the bases on each side of the reaction arrow, identify which side of the equilibrium would be favored in the following hypothetical reactions.

(a)

Textbook Question

For each of the following acid–base reactions, (i) predict which side of the reaction you expect to be favored. If a pK_a is not one of the ten common ones we learned in Chapter 4, it will be given to you.

(a)

Textbook Question

For each of the following reactions, suggest which solvent(s) would be compatible with the acids and bases involved. (We will ignore any other possible reactions for now.) Your choices of solvents are pentane, diethyl ether, ethanol, water, and ammonia. Refer to Appendix 4 for any needed values of pK_a, or estimate them.

Textbook Question

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.

1. CH₃CH₂COOH (pK_a= 4.9)

2. CH₃N⁺H₃ (pK_a= 10.7)

Multiple Choice

In the context of acid-base equilibrium, which of the following factors can disrupt the Hardy-Weinberg equilibrium?

Multiple Choice

Which of the following statements best explains why scientists study acid-base equilibrium in organic chemistry?

Show more practices