Question

Calculate the solubility of CaF2 in water at . You'll find data in the ALEKS Data tab. Round your answer to significant digits.

Answer 1

Answer:

$0.016\frac{g}{L}$

Explanation:

Hello,

In this case, the dissociation of calcium fluoride is:

$CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq)$

And the equilibrium expression is:

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

Which is useful to compute the molar solubility, symbolized by $x$ as the reaction extent:

$Ksp=(x)(2x)^2$

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:

$3.45x10^{-11}=(x)(2x)^2\\\\x=\sqrt[3]{\frac{3.45x10^{-11}}{4} }\\ \\x=2.05x10^{-4}M=2.05x10^{-4}\frac{molCaF_2}{L}$

And the solubility, considering its molar mass 78.08 g/mol is:

$=2.05x10^{-4}\frac{molCaF_2}{L}*\frac{78.08gCaF_2}{L}\\ \\=0.016\frac{g}{L}$

Regards.