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.

S = 0.788 g/L

Explanation:

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

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 is necessary 2 mols of , so if we call "x" the molar concentration of , for we will have 2x, so:

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

m=9

step-by-step explanation:

the ratio of the conjugate base to the acid required to attain the desired ph.