Multiple Choice
Ksp = 2 × 10^(-19) for LaF3. Calculate the solubility of LaF3 in grams per liter in a solution that is 0.013 M in KF.
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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The molar solubility of Ca(OH)₂ was experimentally determined to be 0.020 M. Based on this value, what is the Ksp of Ca(OH)₂?
A
4.0 x 10⁻⁴
B
1.6 x 10⁻⁵
C
8.0 x 10⁻⁶
D
2.0 x 10⁻³
Verified step by step guidance1
Start by writing the balanced chemical equation for the dissolution of calcium hydroxide in water: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq).
Identify the stoichiometry of the dissolution process. For every mole of Ca(OH)₂ that dissolves, 1 mole of Ca²⁺ and 2 moles of OH⁻ are produced.
Express the molar solubility of Ca(OH)₂ in terms of the concentrations of the ions. If the molar solubility is 0.020 M, then [Ca²⁺] = 0.020 M and [OH⁻] = 2 × 0.020 M = 0.040 M.
Write the expression for the solubility product constant (Ksp) for Ca(OH)₂: Ksp = [Ca²⁺][OH⁻]².
Substitute the concentrations of the ions into the Ksp expression: Ksp = (0.020 M) × (0.040 M)². Calculate this expression to find the Ksp value.
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