The solubility product

Let $x$ $\frac{mol}{L}$ be the number of moles of calcium fluoride dissolved in $1$ L of a $0.01$ M solution of $NaF$ at $20^oC$. $x$ is the "solubility" of calcium fluoride in this solution. The dissolution of calcium fluoride $CaF_2(s)$ $Ca^{2+}(aq)$ + $2F^-(aq)$ $K_s=1.7\cdot 10^{-10}\frac{mol}{L}$ supplies per L .......... mol $Ca^{2+}$ and .......... mol $F^-$ Sodium fluoride contributes per liter with .......... mol $F^-$ , in such a way that there is finally in the saturated solution: $[Ca^{2+}]$ = ..........$\frac{mol}{L}$ $[F^-]$ = ..........$\frac{mol}{L}$ The product of solubility at $25^oC $ of $CaF_2$, expressed as a function of $x$, is therefore worth $K_s=[Ca^{2+}][F^-]^2$ = .......... $\frac{mol^3}{L^3}$ - Establish the equation giving the solubility of calcium fluoride in $NaF$ $0.01$ M using these data! - Solve with an efficient computing machine

Calculation: ....... $x(2x$ $+$ $0,01)^2$ $=$ $K_s$ $x(2x$ $+$ $0,01)^2$ $=$ $1,7\cdot 10^{-10}$ ....... $x$ $=$ $1.7\cdot 10^{-6}$ ....... Solubility of $CaF_2$ in $ NaF \; 0.01 M$ $ =$ $ 1.7\cdot 10^{-6}\frac{mol}{L}$