Question 12.EP.11: What is the molar solubility of calcium phosphate, given tha...

The relevant equilibrium is the dissolution of calcium phosphate:

Ca_{3}(PO_{4})_{2}(s)\rightleftarrows 3 \ Ca^{2+}(aq)+2 \ PO_{4}^{ \ 3-}(aq)

We will need to work with the solubility product expression for this reaction, and the value of K_{sp} is given.

K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{ \ 3-}]^{2}=2.0\times 10^{-33}

We can envision the process as starting with only solid present. Dissolution of some of the solid leads to the presence of ions in the saturated solution at equilibrium. In setting up our table, we must be sure to account for the 3:2 stoichiometry. For every mole of Ca_{3}(PO_{4})_{2}(s) that dissolves, three moles of calcium ion and two moles of phosphate ion are released into solution.

Substitute in the equilibrium expression:

K_{sp} =2.0\times 10^{-33}=[Ca^{2+}]^{3}[PO_{4}^{ \ 3-}]^{2}=(3x)^{3}(2x)^{2}

We can solve this for x:

2.0 × 10^{−33} = 108x^{5}

x = 1.1 × 10^{−7}

This means that the molar solubility of calcium phosphate is 1.1 × 10^{−7}M. Now convert units to find the solubility in terms of mass.

Solubility = \frac{1.1 × 10^{−7} \ mol \ Ca_{3}(PO4)_{2}}{L}\times \frac{1 \ L}{10^{3}cm^{3}} \times \frac{1 \ cm^{3}}{1.0 \ g}\times \frac{310.2 \ g \ Ca_{3}(PO4)_{2}}{1 \ mol \ Ca_{3}(PO4)_{2}}

\               = 3.5 × 10^{−8} g solute/g H_{2}O

Multiplying this by 100 gives us a solubility of 3.5 × 10^{−6} g Ca_{3}(PO4)_{2} per 100 g H_{2}O.

Analyze Your Answer
The molar solubility we obtained is very small. Does that make sense? The value for K_{sp} is very small, so we can expect that the equilibrium will favor reactants strongly. Favoring reactants means few ions in solution, so a small solubility makes sense.