Respuesta :

PBCHEM
Following is the dissociation reaction of CrF3

CrF3              →            Cr3+            +             3F-

Now, solubility product (Ksp) = [Cr3+] [F-]^3

Let [Cr3+]  =   [F-]   =   x

∴Ksp = x(x)^3 
         = x^4

But Given: Ksp = 6.6 X 10^-11

6.6 X 10^-11 = x^4
x = 2.85 x 10^-3 M

Thus, the molar solubility of chromium(iii) fluoride is 2.83 X 10^-3 M
sebassandin

Answer:

[tex]x=1.3x10^{-3}M[/tex]

Explanation:

Hello,

in this case, one considers the dissolution as an ionic reaction:

[tex]CrF_3<-->Cr^{+3}+3F^{-}[/tex]

In such a way, we write the equilibrium equation based on the change [tex]x[/tex] due to the slight rate of dissolution of the considered salt as:

[tex]Ksp=[Cr^{+3} ]_{eq} [F^{-} ]_{eq} ^{3} \\Ksp=x(3x)^{3}  \\x=\sqrt[4]{\frac{Ksp}{27}} =\sqrt[4]{\frac{6.6x10^{-11} }{27}}\\x=1.3x10^{-3}M[/tex]

Such [tex]x[/tex] accounts for the molar solubility of the chromium (III) fluoride.

Best regards.