Question

Example: One liter of saturated calcium fluoride

solution contains 0.0167 gram of CaFat 25°C.
Calculate the molar solubility of, and Ksp for, CaF2.​

Answer 1

`Molar\ solubility=2.14x10^(-4)M`

`Ksp=3.91x10^(-11)`

Step-by-step 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=(0.0167gCaF_2)/(1L)*(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.