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?
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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Calculate the molar solubility of calcium fluoride (CaF2) in g/L if its solubility product constant (Ksp) is 5.3×10^-11.
A
0.0016 g/L
B
0.16 g/L
C
0.016 g/L
D
0.001 g/L
Verified step by step guidance1
Start by writing the balanced chemical equation for the dissolution of calcium fluoride (CaF2) in water: CaF2(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
Express the solubility product constant (Ksp) for CaF2. The Ksp expression is: .
Let the molar solubility of CaF2 be 's'. Therefore, the concentration of Ca²⁺ ions will be 's' and the concentration of F⁻ ions will be '2s'. Substitute these into the Ksp expression: .
Set the Ksp value equal to the expression derived: . Solve for 's' to find the molar solubility in mol/L.
Convert the molar solubility from mol/L to g/L using the molar mass of CaF2. The molar mass of CaF2 is approximately 78.08 g/mol. Use the formula: to find the solubility in g/L.
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