The K_sp of Ca₃(PO₄)₂ is 2.0 × 10^–33. Determine its molar solubility in 0.05 M (NH₄)₃PO₄. Compare your answer to the molar solubility of Ca3(PO4)2 in water, which we calculated in Example Problem 12.11.

Strategy (NH₄)₃PO₄ is reasonably soluble in water, so [PO₄ ^3–] in 0.05 M (NH₄)₃PO₄ is 0.05 M. When solid Ca₃(PO₄)₂ is added and equilibrium is established, molar solubility can be found from the amount of [Ca²⁺] in solution. Write the reaction of interest and set up the usual equilibrium table. Because the value of K_sp is very small, we can use a simple approximation to solve for molar solubility.

Example Problem 12.11

What is the molar solubility of calcium phosphate, given that K_sp = 2.0 × 10^–33?

Assuming that the density of the saturated solution is 1.00 g/cm³, what is the solubility in grams of Ca₃(PO₄)₂ per 100 grams of solvent?

Strategy This problem asks us to find the equilibrium concentration of ions in a saturated solution. So it is actually similar to problems we solved earlier in this chapter. Hence we will use the same general approach. To find molar solubility, begin by creating an equilibrium table for a solution with unknown equilibrium concentrations and solve the resulting equilibrium expression for the unknown. Convert the concentration units to grams of solute per 100 grams of solvent, using the molar mass of the solute and the density of the solution. (Because the solution will be very dilute, we will assume that there is no significant difference between 100 g of solution and 100 g of solvent.)