Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

17. Acid and Base Equilibrium

pH of Weak Bases

General Chemistry

17. Acid and Base Equilibrium

pH of Weak Bases

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Multiple Choice

A solution of sodium acetate (NaCH3COO) has a pH of 9.64. The acid-dissociation constant (Ka) for acetic acid is 1.8 x 10^-5. What is the molarity of the sodium acetate solution?

0.10 M

0.25 M

1.00 M

0.50 M

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Verified step by step guidance

Start by understanding that sodium acetate (NaCH3COO) is a salt derived from acetic acid (CH3COOH) and sodium hydroxide (NaOH). In solution, it can hydrolyze to form acetic acid and hydroxide ions, affecting the pH.

Use the given pH of the solution, 9.64, to find the concentration of hydroxide ions [OH^-]. The relationship between pH and pOH is given by: \( \text{pOH} = 14 - \text{pH} \). Calculate pOH and then find [OH^-] using \( [\text{OH}^-] = 10^{-\text{pOH}} \).

Recognize that the hydrolysis of sodium acetate in water can be represented by the equilibrium: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \). The equilibrium constant for this reaction, Kb, can be found using the relation \( K_b = \frac{K_w}{K_a} \), where \( K_w = 1.0 \times 10^{-14} \) is the ion-product constant for water.

Set up an expression for Kb using the concentrations at equilibrium: \( K_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \). Assume that the initial concentration of sodium acetate is \( C \) and that the change in concentration due to hydrolysis is \( x \), where \( x = [\text{OH}^-] \).

Solve for \( C \) using the expression \( K_b = \frac{x^2}{C - x} \). Substitute the values for \( K_b \) and \( x \) obtained from previous steps, and solve for \( C \), which represents the molarity of the sodium acetate solution.