Question

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

FeF2(s) ⇌ Fe^2+(aq) + 2 F^–(aq)

Calculate the solubility of FeF2 in water by using 2.36 x 10^-6 as the solubility product.

Answer 1

C. Is the correct answer

Step-by-step explanation:

Answer 2

S = 0.788 g/L

Step-by-step explanation:

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

`Kps = ([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