The quantity of fluoride ions $(F^{-})$ at $25^oC$ in a saturated solution of pure strontium fluoride is equal to $0.0220$ g in $100 $ mL .
The molarity of the fluoride in the saturated solution is therefore equal to .......... $\frac{mol}{L}$
The equation of dissolution of pure strontium fluoride
$SrF_2(s)$ 
$Sr^{2+}(aq)$ $+$ $2F^{-}(aq)$
shows that in that case the concentration of $Sr^{2+}$ is equal to .......... $\frac{mol}{L}$
The solubility product at $15^oC$ of $[SrF_2]$ is therefore equal to $K_s$ $=$ $[Sr^{2+}][F^{-}]^2$ = .......... $\frac{mol^3}{L^3}$
Click where you want to see the answer! Finish this question before beginning the next one!