Answer :
The given question is incomplete. The complete question is as follows.
The solubility product of calcium fluoride () is at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
Explanation:
Reaction equation for the given chemical reaction is as follows.
[tex]CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}[/tex]
Therefore, expression for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp} = [Ca^{2+}][F^{-}]^{2}[/tex]
=
Also, moles of per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]
= [tex]\frac{1.0 \times 10^{-3}}{19.0}[/tex]
= [tex]5.263 \times 10^{-5} mol[/tex]
Hence, [tex][F^{-}] = \frac{\text{moles of F^{-}}{volume}[/tex]
= [tex]\frac{5.263 \times 10^{-5}}{1}[/tex]
= M
Now, moles of per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]
= [tex]\frac{200 \times 10^{-3}}{40.1}[/tex]
= M
Also, [tex][Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}[/tex]
= [tex]\frac{4.988 \times 10^{-3}}{1}[/tex]
= M
Hence, ionic product =
= [tex](4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}[/tex]
= [tex]1.38 \times 10^{-11}[/tex]
As, the ionic product is less than the [tex]K_{sp}[/tex], this means that the fluoride will be soluble in water containing the calcium.