The solubility product

The maximum amount of lead fluoride that dissolves in 1 L of water at $20^oC$ is $1,9 \cdot 10^{-3}$ mol. The .......... of $PbF_2$ (not to be confused with the product of solubility!) is therefore worth $1,9 \cdot 10^{-3}\frac{mol}{L}$ This dissolution is immediately followed by dissociation into ions: $PbF_2(s)$ $Pb^{2+}(aq)$ $+$ $2F^{-}(aq)$ In $1$ L of a .......... solution of $PbF_2$ in equilibrium with solid undissolved $PbF_2$ there is .......... mol $[Pb^{2+}]$ et .......... mol $[F^-]$. So, the solubility product at $20^oC$ of $[PbF_2]$ is $K_s=[Pb^{2+}][F^-]^2$ = .......... $\frac{mol^3}{L^3}$