Answer:
S = 0.788 g/L
Explanation:
The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:
[tex]Kps = \frac{[product]^x}{[reagent]^y}[/tex]
Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.
Analyzing the equation, we see that for 1 mol of [tex] Fe^{+2}[/tex] is necessary 2 mols of [tex]F^-[/tex], so if we call "x" the molar concentration of [tex] Fe^2[/tex], for [tex]F^-[/tex] we will have 2x, so:
[tex]Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L[/tex]
So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:
Fe = 55.8 g/mol
F = 19 g/mol
FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol
So,
[tex]S = 8.4x10^{-3}x93.8
S = 0.788 g/L
