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1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 36m
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 0.1 M solution of ammonia (NH3), a weak base with a Kb of 1.8 x 10^-5?

11.13

9.25

12.75

7.00

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Verified step by step guidance

Identify that ammonia (NH3) is a weak base and will partially ionize in water to form NH4+ and OH-. The equilibrium expression for this reaction is: NH3 + H2O ⇌ NH4+ + OH-.

Write the expression for the base dissociation constant (Kb) for ammonia: Kb = [NH4+][OH-] / [NH3].

Assume that the initial concentration of NH3 is 0.1 M and that the change in concentration due to ionization is 'x'. Therefore, at equilibrium, [NH4+] = x, [OH-] = x, and [NH3] = 0.1 - x.

Substitute these equilibrium concentrations into the Kb expression: 1.8 x 10^-5 = (x)(x) / (0.1 - x). Since Kb is small, assume x << 0.1, simplifying the expression to 1.8 x 10^-5 ≈ x^2 / 0.1.

Solve for x to find the concentration of OH- ions, then calculate the pOH using the formula pOH = -log[OH-]. Finally, use the relationship pH + pOH = 14 to find the pH of the solution.