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

Calculate the pH of a 0.30 M solution of barium nitrite. Given that the Ka of HNO2 is 4.5 x 10^-4, what is the pH of the solution?

5.48

8.52

9.25

7.00

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Verified step by step guidance

Identify that barium nitrite, Ba(NO2)2, is a salt that will dissociate in water to form Ba^2+ and NO2^- ions. The NO2^- ion is the conjugate base of the weak acid HNO2.

Recognize that the NO2^- ion will undergo hydrolysis in water, reacting with water to form HNO2 and OH^- ions. This reaction can be represented as: NO2^- + H2O ⇌ HNO2 + OH^-.

Use the given Ka value of HNO2 to find the Kb for NO2^- using the relation Kb = Kw / Ka, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C).

Set up an expression for the equilibrium concentrations using the Kb expression: Kb = [HNO2][OH^-] / [NO2^-]. Assume that the initial concentration of NO2^- is 0.30 M and that the change in concentration due to hydrolysis is 'x'.

Solve for 'x', which represents the concentration of OH^- ions, and then calculate the pOH of the solution. Finally, use the relation pH + pOH = 14 to find the pH of the solution.