Multiple Choice
Calculate the molar solubility of CuX in a 0.18 M Na2X solution, given that the solubility product constant (Ksp) for CuX is 1.27×10^-36.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
What is the molar solubility of barium fluoride (BaF₂) in a 0.12 M Ba(NO₃)₂ solution, given that the solubility product constant (Ksp) for BaF₂ is 2.45 × 10⁻⁵?
A
2.45 × 10⁻⁵ M
B
1.23 × 10⁻⁴ M
C
1.02 × 10⁻⁵ M
D
3.67 × 10⁻⁶ M
Verified step by step guidance1
Understand the concept of molar solubility and the solubility product constant (Ksp). Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to reach saturation. The Ksp is the equilibrium constant for the dissolution of a sparingly soluble compound.
Write the balanced chemical equation for the dissolution of barium fluoride (BaF₂) in water: BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).
Express the Ksp in terms of the concentrations of the ions produced: Ksp = [Ba²⁺][F⁻]². Since BaF₂ dissociates into one Ba²⁺ ion and two F⁻ ions, if 's' is the molar solubility of BaF₂, then [Ba²⁺] = s and [F⁻] = 2s.
Consider the common ion effect due to the presence of Ba(NO₃)₂ in the solution. The concentration of Ba²⁺ from Ba(NO₃)₂ is 0.12 M, which affects the equilibrium. Therefore, the total [Ba²⁺] = 0.12 + s.
Substitute the expressions for [Ba²⁺] and [F⁻] into the Ksp expression: Ksp = (0.12 + s)(2s)². Solve this equation for 's', the molar solubility of BaF₂ in the solution.
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