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

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

General Chemistry

18. Aqueous Equilibrium

Solubility Product Constant: Ksp

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Multiple Choice

Determine the molar solubility of Al(OH)₃ in a solution containing 0.05 M AlCl₃. Ksp of Al(OH)₃ = 1.3 x 10⁻³³.

1.0 x 10⁻³⁴ M

5.2 x 10⁻³² M

2.6 x 10⁻³² M

1.3 x 10⁻³³ M

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Verified step by step guidance

Understand the concept of molar solubility and the common ion effect. Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to reach saturation. The common ion effect occurs when a salt's solubility is reduced due to the presence of a common ion in the solution.

Write the dissolution equation for Al(OH)₃: \( \text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3\text{OH}^- (aq) \). This equation shows that Al(OH)₃ dissociates into Al³⁺ and OH⁻ ions in solution.

Use the solubility product constant (Ksp) expression for Al(OH)₃: \( K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \). Substitute the known value of Ksp, which is \( 1.3 \times 10^{-33} \).

Consider the initial concentration of Al³⁺ from AlCl₃, which is 0.05 M. This concentration affects the equilibrium concentration of Al³⁺ due to the common ion effect. Let the molar solubility of Al(OH)₃ be \( s \). Therefore, the concentration of Al³⁺ at equilibrium is \( 0.05 + s \), and the concentration of OH⁻ is \( 3s \).

Substitute these expressions into the Ksp equation: \( 1.3 \times 10^{-33} = (0.05 + s)(3s)^3 \). Solve this equation for \( s \), the molar solubility of Al(OH)₃ in the solution.