Multiple Choice
Ca3(PO4)2 is being dissolved in water. If only 0.46 g of Ca3(PO4)2 will dissolve in 100 mL of water at 25°C, what is the solubility product, Ksp, of this salt at this temperature?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
What is the molar solubility of barium fluoride (BaF₂) in pure water if the solubility product constant (Ksp) is 2.45×10⁻⁵?
A
5.00×10⁻³ M
B
2.45×10⁻⁵ M
C
3.11×10⁻³ M
D
1.24×10⁻² M
Verified step by step guidance1
Identify the dissolution equation for barium fluoride (BaF₂) in water: BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).
Define the molar solubility of BaF₂ as 's'. This means that at equilibrium, the concentration of Ba²⁺ ions will be 's' and the concentration of F⁻ ions will be '2s'.
Write the expression for the solubility product constant (Ksp) for BaF₂: Ksp = [Ba²⁺][F⁻]².
Substitute the equilibrium concentrations into the Ksp expression: Ksp = (s)(2s)² = 4s³.
Set the Ksp expression equal to the given Ksp value and solve for 's': 4s³ = 2.45×10⁻⁵. This will give you the molar solubility of BaF₂ in pure water.
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