Question

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

Answer 1

`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?