Question

If the solubility of calcium phosphate
(molecular weight M) in water at 25°C is w g
per 100 mL, its solubility product at 25°C is
approximately :​

Answer 1

Answer:

The solubility product, by definition, is the equilibrium constant for dissolution of a solid ionic compound to yield ions in solution (usually water). It is abbreviated Ksp.

The equation for dissolution of calcium phosphate (Ca3(PO4)2) in water is:

Ca3(PO4)2 = 3 Ca(2+) + 2 PO4(3-)

and the equilibrium constant will be equal to:

Ksp = [Ca(2+)]^3 * [PO4(3-)]^2

Now, if you want to solve problems like how much calcium phosphate will dissolve in water you need to replace [Ca(2+)] and [PO4(3-)] with something more useful. For every mole of Ca3(PO4)2 (set = to x), when it dissolves, will yield 3 moles of Ca(2+) (=3x) and 2 moles of PO4(3-) (=2x) so the Ksp expression becomes:

Ksp = [Ca(2+)]^3 * [PO4(3-)]^2 => (3x)^3 * (2x)^2 = 108x^5