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.

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:

$K_{sp}=3.4\times 10^{-11}$

Putting values in above equation, we get:

$3.4\times 10^{-11}=(s)\times (2s)^2\\\\3.4\times 10^{-11}=4s^3\\\\s=2.04\times 10^{-4}mol/L$

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:

$s=2.04\times 10^{-4}mol/L\times 78g/mol=159.12\times 10^{-4}g/L=0.015g/L$

Hence, the correct answer is Option b.