Given that the solubility of #BaF_2# is #4.59xx10^(-2)*mol*L^(-1)# under standard conditions, what is #K_"sp"# for barium fluoride?

1 Answer
Mar 20, 2017

Answer:

#K_"sp"=4.59xx10^-2*mol*L^-1#

Explanation:

We interrogate the equilibrium:

#BaF_2(s) rightleftharpoonsBa^(2+) + 2F^-#

Now #K_"sp"=([Ba^(2+)][F^-]^2)/([BaF_2(s)])#

Now #[BaF_2(s)]#, as a solid, is UNDEFINED, and treated as unity.

So #K_"sp"=[Ba^(2+)][F^-]^2=??#

But we are GIVEN that #[BaF_2]=4.59xx10^-2*mol*L^-1#.

And thus #[Ba^(2+)]=4.59xx10^-2*mol*L^-1#

And #[F^-]=2xx4.59xx10^-2*mol*L^-1#

And so #K_"sp"=(4.59xx10^-2)(2xx4.59xx10^-2)^2#

#=4xx(4.59xx10^-2)^3=3.87xx10^-4#.

This site reports that #K_"sp"# of #"barium fluoride"# is #1.84xx10^-7# at #25# #""^@C#. Given the #K_"sp"# calculated here, which did NOT specify the conditions, would this be at a temperature higher or lower than #298*K#? Why?