Answer: The correct answer is Option b.
Explanation:
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
[tex]CaF_2\rightleftharpoons Ca^{2+}+2F^-[/tex]
s 2s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ca^{2+}][F^-]^2[/tex]
We are given:
[tex]K_{sp}=3.4\times 10^{-11}[/tex]
Putting values in above equation, we get:
[tex]3.4\times 10^{-11}=(s)\times (2s)^2\\\\3.4\times 10^{-11}=4s^3\\\\s=2.04\times 10^{-4}mol/L[/tex]
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:
[tex]s=2.04\times 10^{-4}mol/L\times 78g/mol=159.12\times 10^{-4}g/L=0.015g/L[/tex]
Hence, the correct answer is Option b.
