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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

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

General Chemistry

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

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Multiple Choice

What is the molar solubility of barium fluoride (BaF₂) in pure water if the solubility product constant (Ksp) is 2.45×10⁻⁵?

5.00×10⁻³ M

2.45×10⁻⁵ M

3.11×10⁻³ M

1.24×10⁻² M

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Verified step by step guidance

Identify the dissolution equation for barium fluoride (BaF₂) in water: BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).

Define the molar solubility of BaF₂ as 's'. This means that at equilibrium, the concentration of Ba²⁺ ions will be 's' and the concentration of F⁻ ions will be '2s'.

Write the expression for the solubility product constant (Ksp) for BaF₂: Ksp = [Ba²⁺][F⁻]².

Substitute the equilibrium concentrations into the Ksp expression: Ksp = (s)(2s)² = 4s³.

Set the Ksp expression equal to the given Ksp value and solve for 's': 4s³ = 2.45×10⁻⁵. This will give you the molar solubility of BaF₂ in pure water.