The solubility product

At $20^oC$, calcium fluoride ($CaF_2$) has the following solubility product $K_s$ $=$ $1.7\cdot10^{-10}\frac{mol^3}{L^3}$ Let $x$ $\frac{mol}{L}$ be the solubility of calcium fluoride. The dissolution of the pure salt is given by: $CaF_2(s)$ $Ca^{2+}(aq)$ $+$ $2F^{-}(aq)$ Then, expressed as a function of $x$, the molarity of the fluoride ion in the saturated solution is equal to .......(2) The product of solubility at $20^oC$ of $CaF_2$, expressed as a function of $x$, is therefore worth $K_s$ $=$ $[Ca^{2+}][F^{-}]^2$ = ........(1) $\frac{mol^3}{L^3}$ Calculate the solubility using this data!

Calculation: .......(2) $4x^3$ $=$ $K_s$ $4x^3$ $=$ $1.7\cdot 10^{-10}$ .......(3) $x$ $=$ $(\frac{1.7\cdot 10^{-10}}{4})^{\frac{1}{3}}$ $=$ $3.5\cdot10^{-4}$ .......(4) Solubility of calcium fluoride = $3.5\cdot10^{-4}\frac{mol}{L}$