Multiple Choice
Calculate the molar solubility of MgF2 in 0.15 M MgCl2 at 25 °C, given that the Ksp of MgF2 is 6.4 x 10^-9.
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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Calculate the molar solubility of barium fluoride (BaF₂) in pure water, given that its solubility product constant (Ksp) is 2.45 × 10⁻⁵.
A
2.45 × 10⁻⁵ M
B
1.57 × 10⁻³ M
C
1.24 × 10⁻² M
D
1.57 × 10⁻² M
Verified step by step guidance1
Start by writing the balanced dissolution equation for barium fluoride (BaF₂) in water: BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).
Express the solubility product constant (Ksp) for the dissolution of BaF₂. The Ksp expression is: Ksp = [Ba²⁺][F⁻]².
Let the molar solubility of BaF₂ be 's'. This means [Ba²⁺] = s and [F⁻] = 2s, because for every mole of BaF₂ that dissolves, one mole of Ba²⁺ and two moles of F⁻ are produced.
Substitute the expressions for [Ba²⁺] and [F⁻] 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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