The solubility of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$in solution in $1.6\times {10}^{-7}{\mathrm{molL}}^{-1}$. Its solubility product is

$\begin{array}{l}{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\rightleftharpoons \underset{\left(3s\right)}{2{\mathrm{Ca}}^{2+}}+\underset{\left(2s\right)}{2{\mathrm{PO}}_4^3}\\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }{K}_{\mathrm{s}}={\left[{\mathrm{Ca}}^{2+}\right]}^3{\left[{\mathrm{PO}}_4^{3-}\right]}^2=\left(3s{)}^3\right(2s{)}^2=108s^5=\left(108\right){\left(1.6\times {10}^{-7}\mathrm{M}\right)}^5=1.13\times {10}^{-32}{\mathrm{M}}^5\end{array}$