Question

What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11 at 25°C. a. 2.7 × 10–9 g/L b. 0.015 g/L c. 1.3 × 10–9 g/L d. 0.00045 g/L e. 0.094 g/L

Answer 1

Answer: The correct answer is Option b.

Step-by-step explanation:

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

`CaF_2\rightleftharpoons Ca^(2+)+2F^-`

s 2s

The expression for solubility constant for this reaction will be:

`K_(sp)=[Ca^(2+)][F^-]^2`

We are given:

`K_(sp)=3.4* 10^(-11)`

Putting values in above equation, we get:

`3.4* 10^(-11)=(s)* (2s)^2\\\\3.4* 10^(-11)=4s^3\\\\s=2.04* 10^(-4)mol/L`

To calculate the solubility in g/L, we will multiply the calculated solubility with the molar mass of calcium fluoride:

Molar mass of calcium fluoride = 78 g/mol

Multiplying the solubility product, we get:

`s=2.04* 10^(-4)mol/L* 78g/mol=159.12* 10^(-4)g/L=0.015g/L`

Hence, the correct answer is Option b.