Question

The solubility product of calcium fluoride (CaF2(s); fluorite) is 310-11 at 25C. Could a fluoride concentration of 1.0 mg L-1 be obtained in water that contains 200 mg L-1 of calcium?

Answer 1

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.

       $CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}$

Therefore, expression for $K_{sp}$ will be as follows.

        $K_{sp} = [Ca^{2+}][F^{-}]^{2}$

                     =

Also, moles of  per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]

                = $\frac{1.0 \times 10^{-3}}{19.0}$

               = $5.263 \times 10^{-5} mol$

Hence,    $[F^{-}] = \frac{\text{moles of F^{-}}{volume}$

                       = $\frac{5.263 \times 10^{-5}}{1}$

                      =  M

Now, moles of  per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]

            = $\frac{200 \times 10^{-3}}{40.1}$

           =  M

Also,   $[Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}$

                      = $\frac{4.988 \times 10^{-3}}{1}$

                     =  M

Hence, ionic product =

                 = $(4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}$

                = $1.38 \times 10^{-11}$

As, the ionic product is less than the $K_{sp}$, this means that the fluoride will be soluble in water containing the calcium.