Question 12.EP.12: The Ksp of Ca3(PO4)2 is 2.0 × 10^−33. Determine its molar so...

Let x = moles of Ca_{3}(PO_{4})_{2} that dissolve per liter (i.e., the molar solubility that we want to calculate). The initial concentration of Ca^{2+} is zero, but that of PO_{4}^{ \ 3−} is now 0.10 M. This gives us the following equilibrium table:

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

Substitute in the solubility product expression:

K_{sp} =2.0\times 10^{-33}=[Ca^{2+}]^{3}[PO_{4}^{ \ 3-}]^{2}=(3x)^{3}(0.10+2x)^{2}

This looks daunting: multiplying out would give us a term involving x^{5}. But we can take advantage of a clever approximation. We know from the last example problem that the solubility in pure water was on the order of 10^{-7} M. And in this case the presence of the common ion should decrease the solubility considerably from that level. So we know that x will be less than 10^{-7}. This means that we can neglect the +2x piece of the last term in comparison to 0.10.

0.10 + 2x ≈ 0.10

This simplifies the problem considerably.

2.0\times 10^{-33}=(3x)^{3}(0.10)^{2}

x = 1.9 × 10^{-11} M

In the previous example problem, we calculated the molar solubility of Ca_{3}(PO_{4})_{2} in water as 1.1 × 10^{-7} M. The presence of a high concentration of phosphate ion in solution has depressed the solubility by almost four orders of magnitude, to 1.9 × 10^{-11} M.

Analyze Your Answer
Whenever we make a simplifying approximation of the sort used here, it is important to verify that the eventual solution is consistent with the underlying assumption. Here, we assumed that 2x was negligible compared to 0.10. Because we found that x was on the order of 10^{-11}, that assumption was certainly justifiable. We could have used appropriate software tools to find a numerical solution to the actual equation, but the added work would not have improved our result discernibly. Note that the underlying reason why this assumption could be made was the fact that the K_{sp} value was a very small number.