Question

Example: One liter of saturated calcium fluoride

Answer 1

Answer:

$Molar\ solubility=2.14x10^{-4}M$

$Ksp=3.91x10^{-11}$

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):

$Molar\ solubility=\frac{0.0167gCaF_2}{1L}*\frac{1molCaF_2}{78.1gCaF_2} \\\\Molar\ solubility=2.14x10^{-4}M$

Next, since dissociation equation for calcium fluoride is:

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

The equilibrium expression is:

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

We can compute the solubility product by remembering that the concentration of both calcium and fluoride ions equals the molar solubility, thereby:

$Ksp=(2.14x10^{-4})(2*2.14x10^{-4})^2\\\\Ksp=3.91x10^{-11}$

Regards.