Question

The solubility product for chromium(iii) fluoride is ksp = 6.6 × 10–11. what is the molar solubility of chromium(iii) fluoride?

Answer 1

`x=1.3x10^(-3)M`

Step-by-step explanation:

Hello,

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

`CrF_3<-->Cr^(+3)+3F^(-)`

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

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

Such
`x` accounts for the molar solubility of the chromium (III) fluoride.

Best regards.