Answer:
[tex]Molar\ solubility=2.14x10^{-4}M[/tex]
[tex]Ksp=3.91x10^{-11}[/tex]
Explanation:
Hello,
In this case, given that 0.0167 grams of calcium fluoride in 1 L of solution form a saturated one, we can notice it is the solubility, therefore, the molar solubility is computed by using the molar mass of calcium fluoride (78.1 g/mol):
[tex]Molar\ solubility=\frac{0.0167gCaF_2}{1L}*\frac{1molCaF_2}{78.1gCaF_2} \\\\Molar\ solubility=2.14x10^{-4}M[/tex]
Next, since dissociation equation for calcium fluoride is:
[tex]CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
We can compute the solubility product by remembering that the concentration of both calcium and fluoride ions equals the molar solubility, thereby:
[tex]Ksp=(2.14x10^{-4})(2*2.14x10^{-4})^2\\\\Ksp=3.91x10^{-11}[/tex]
Regards.
