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

`0.016(g)/(L)`

Step-by-step 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]{(3.45x10^(-11))/(4) }\\ \\x=2.05x10^(-4)M=2.05x10^(-4)(molCaF_2)/(L)`

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

`=2.05x10^(-4)(molCaF_2)/(L)*(78.08gCaF_2)/(L)\\ \\=0.016(g)/(L)`

Regards.