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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

Calculate the molar solubility of barium fluoride (BaF2) in pure water, given that its solubility product constant (Ksp) is 2.45 x 10^-5.

4.9 x 10^-5 M

1.7 x 10^-2 M

3.1 x 10^-4 M

2.5 x 10^-3 M

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Verified step by step guidance

Start by writing the balanced chemical equation for the dissolution of barium fluoride (BaF2) in water: BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).

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

Let the molar solubility of BaF2 be 's'. When BaF2 dissolves, it produces 's' moles of Ba²⁺ and '2s' moles of F⁻. Substitute these into the Ksp expression: Ksp = (s)(2s)².

Simplify the expression: Ksp = 4s³. This equation relates the solubility product constant to the molar solubility.

Solve for 's' by substituting the given Ksp value (2.45 x 10^-5) into the equation: 4s³ = 2.45 x 10^-5. Rearrange to find s: s = (2.45 x 10^-5 / 4)^(1/3).