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

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, CH3NH2, with 25.00 mL of 0.10 M methylammonium chloride, CH3NH3Cl? Assume that the volume of the solutions are additive and that Kb = 3.70 × 10⁻⁴ for methylamine.

9.25

10.61

4.39

7.00

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Verified step by step guidance

Identify that the solution is a buffer solution, consisting of a weak base (methylamine, CH3NH2) and its conjugate acid (methylammonium chloride, CH3NH3Cl).

Use the Henderson-Hasselbalch equation for a buffer solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \).

Calculate \( \text{pK}_a \) from \( K_b \) using the relation \( \text{pK}_a = 14 - \text{pK}_b \). First, find \( \text{pK}_b \) by taking the negative logarithm of \( K_b \): \( \text{pK}_b = -\log(3.70 \times 10^{-4}) \).

Determine the concentrations of the base and acid in the final solution. Since the volumes are additive, the total volume is 50.00 mL. The concentration of each component is halved: \( [\text{CH}_3\text{NH}_2] = \frac{0.10 \text{ M} \times 25.00 \text{ mL}}{50.00 \text{ mL}} \) and \( [\text{CH}_3\text{NH}_3\text{Cl}] = \frac{0.10 \text{ M} \times 25.00 \text{ mL}}{50.00 \text{ mL}} \).

Substitute the values into the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3\text{Cl}]} \right) \). Since the concentrations of the base and acid are equal, the log term becomes zero, simplifying the calculation.