Multiple Choice
Determine the molar solubility of Fe(OH)₂ in pure water given that the Ksp for Fe(OH)₂ is 4.87 × 10⁻¹⁷.
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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If the molar solubility of CaF2 at 35°C is 1.24×10⁻³ mol/L, what is the Ksp at this temperature?
A
1.53×10⁻⁴
B
1.24×10⁻³
C
3.80×10⁻⁹
D
6.10×10⁻⁶
Verified step by step guidance1
Identify the dissolution equation for calcium fluoride (CaF₂) in water: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
Recognize that the molar solubility of CaF₂ is given as 1.24×10⁻³ mol/L, which represents the concentration of Ca²⁺ ions in the solution at equilibrium.
Determine the concentration of fluoride ions (F⁻) in the solution. Since the dissolution of one mole of CaF₂ produces two moles of F⁻, the concentration of F⁻ ions will be 2 × 1.24×10⁻³ mol/L.
Write the expression for the solubility product constant (Ksp) for CaF₂: Ksp = [Ca²⁺][F⁻]².
Substitute the concentrations of Ca²⁺ and F⁻ into the Ksp expression: Ksp = (1.24×10⁻³) × (2 × 1.24×10⁻³)², and simplify to find the Ksp value.
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