The molar solubility of Ba₃(PO₄)₂ is 8.89 × 10^-9M in pure water.Calculate the K_sp for Ba₃(PO₄)₂. A) 4.94 × 10^-49 B) 5.55 × 10^-41 C) 5.33 × 10^-37 D) 8.16 × 10^-31 E) 6.00 × 10^-39

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The dissolution of Ba₃(PO₄)₂ in water can be represented as: Ba₃(PO₄)₂ ⇌ 3Ba²⁺ + 2PO₄³⁻. If the molar solubility is s = 8.89 × 10⁻⁹ M, then [Ba²⁺] = 3s and [PO₄³⁻] = 2s. The Ksp expression is Ksp = [Ba²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 108s⁵. Substituting s = 8.89 × 10⁻⁹ M gives Ksp = 108(8.89 × 10⁻⁹)⁵ = 6.00 × 10⁻³⁹.

The Molar Solubility of Ba3(PO4)2 Is 8

The Molar Solubility of Ba₃(PO₄)₂ Is 8