The solubility of scandium (III) fluoride, #ScF_3#, in pure water is #2.0 x 10^-5# moles per liter. How do you calculate the value of #K_(sp)# for scandium (III) fluoride from this data?

1 Answer
anor277
Sep 12, 2017

#K_"sp"(ScF_3)=4.32xx10^-18#

We interrogate the reaction.....

Explanation:

We interrogate the reaction.....

#ScF_3(s) rightleftharpoons Sc^(3+) + 3F^-#

And as written, we can directly write the solubility expression, #K_"sp"#, where #K_"sp"=[Sc^(3+)][F^-]^3#

And we are given the solubility of #ScF_3# in water, #2.0xx10^-5*mol*L^-1#....

And thus #[Sc^(3+)]=2.0xx10^-5*mol*L^-1#

And #[F^-]=3xx2.0xx10^-5*mol*L^-1#

Capisce?

And then we input these numbers into the #K_"sp"# expression.....

#K_"sp"=(2.0xx10^-5)xx(6.0xx10^-5)^3=4.32xx10^-18#

PS I have not got a good text at hand, but the values I read for #K_"sp"# for scandium fluoride seem to vary widely. I would check the source of these data.

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