Answer:
[tex]0.016\frac{g}{L}[/tex]
Explanation:
Hello,
In this case, the dissociation of calcium fluoride is:
[tex]CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq)[/tex]
And the equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
Which is useful to compute the molar solubility, symbolized by [tex]x[/tex] as the reaction extent:
[tex]Ksp=(x)(2x)^2[/tex]
In such a way, since the solubility product of calcium fluoride at 25 °C is 3.45x10⁻¹¹, the molar solubility is found to be:
[tex]3.45x10^{-11}=(x)(2x)^2\\\\x=\sqrt[3]{\frac{3.45x10^{-11}}{4} }\\ \\x=2.05x10^{-4}M=2.05x10^{-4}\frac{molCaF_2}{L}[/tex]
And the solubility, considering its molar mass 78.08 g/mol is:
[tex]=2.05x10^{-4}\frac{molCaF_2}{L}*\frac{78.08gCaF_2}{L}\\ \\=0.016\frac{g}{L}[/tex]
Regards.
