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 molarity of ions produced in solution?

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The solubility product (Ksp) of Ca3(PO4)2 is given as 6.47×10^-32. The dissolution of Ca3(PO4)2 in water can be represented by the equation: Ca3(PO4)2(s) ⇌ 3Ca^2+(aq) + 2PO4^3-(aq). For every mole of Ca3(PO4)2 that dissolves, 3 moles of Ca^2+ and 2 moles of PO4^3- ions are produced. Let the solubility of Ca3(PO4)2 be S mol/L. Then, [Ca^2+] = 3S and [PO4^3-] = 2S. Substituting these into the expression for Ksp gives: Ksp = [Ca^2+]^3[PO4^3-]^2 = (3S)^3(2S)^2 = 108S^5. Solving for S with Ksp = 6.47×10^-32 gives S ≈ 2.27×10^-7 M, which corresponds to the solubility of Ca3(PO4)2 in terms of the molarity of ions produced in solution.

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

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