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

Answer: (B)

Verified step by step guidance

1. Start by writing the balanced dissolution equation for barium fluoride (BaF₂) in water: BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq).
2. Express the solubility product constant (Ksp) for the dissolution of BaF₂. The Ksp expression is: Ksp = [Ba²⁺][F⁻]².
3. 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.
4. Substitute the expressions for [Ba²⁺] and [F⁻] into the Ksp expression: Ksp = (s)(2s)² = 4s³.
5. 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.