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What is the molar solubility of CaCO3 in ocean water at pH 8.3, given that the solubility is pH dependent and the dissociation constants are Ka1(H2CO3) = 4.3x10^-7 and Ka2(H2CO3) = 5.6x10^-11?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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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.
A
1.0 x 10⁻³⁴ M
B
5.2 x 10⁻³² M
C
2.6 x 10⁻³² M
D
1.3 x 10⁻³³ M
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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.
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