Ca3(PO4)2 is being dissolved in water. If only 0.46 g of Ca3(PO4)2 will dissolve in 100 mL of water at 25 °C, what is the solubility product, Ksp, of this salt at this temperature?

• A. 3.00 x 10^-18
• B. 2.07 x 10^-33
• C. 5.00 x 10^-10
• D. 1.00 x 10^-25

Answer: (B)

Verified step by step guidance

1. First, write the balanced dissolution equation for calcium phosphate, Ca3(PO4)2, in water: Ca3(PO4)2(s) ⇌ 3Ca^2+(aq) + 2PO4^3-(aq).
2. Determine the molar mass of Ca3(PO4)2 by adding the atomic masses of all atoms in the formula: 3(Ca) + 2(P) + 8(O).
3. Calculate the number of moles of Ca3(PO4)2 that dissolve in 100 mL of water using the given mass (0.46 g) and the molar mass from the previous step.
4. Use the stoichiometry of the dissolution reaction to find the concentrations of Ca^2+ and PO4^3- ions in the solution. For every mole of Ca3(PO4)2 that dissolves, 3 moles of Ca^2+ and 2 moles of PO4^3- are produced.
5. Substitute the concentrations of the ions into the expression for the solubility product, Ksp = [Ca^2+]^3[PO4^3-]^2, and solve for Ksp.