If the K_spfor Ca₃(PO₄)₂ at $30 ^ { \circ } \mathrm { C }$ is 6.47×10^-32, what is the aqueous solubility of Ca₃(PO₄)₂in terms of the mass of salt that dissolves in 100mL of water at the same temperature?

Question

Quizplus · Accepted Answer

The solubility product expression for Ca₃(PO₄)₂ is Ksp = [Ca²⁺]³[PO₄³⁻]². Let the solubility be s mol/L. Then, [Ca²⁺] = 3s and [PO₄³⁻] = 2s. Substituting these into the Ksp expression gives Ksp = (3s)³(2s)² = 108s⁵. Solving for s using the given Ksp value, we find s ≈ 1.43×10⁻⁷ mol/L. The molar mass of Ca₃(PO₄)₂ is approximately 310.18 g/mol, so the solubility in g/L is 1.43×10⁻⁷ mol/L × 310.18 g/mol ≈ 4.43×10⁻⁵ g/L. For 100 mL, it is 4.43×10⁻⁶ g, which is closest to option D.

If the Ksp for Ca3(PO4)2 At 30∘C30 ^ { \circ } \mathrm { C }30∘C

If the K_spfor Ca₃(PO₄)₂ At $30 ^ { \circ } \mathrm { C }$