Question

Iron fluoride (FeF2) dissociates according to the following equation:

Answer 1

Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

$Kps = \frac{[product]^x}{[reagent]^y}$

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 $Fe^{+2}$ is necessary 2 mols of $F^-$, so if we call "x" the molar concentration of $Fe^2$, for $F^-$ we will have 2x, so:

$Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L$

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