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

Mar 20, 2017

${K}_{\text{sp}} = 4.59 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

Explanation:

We interrogate the equilibrium:

$B a {F}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s B {a}^{2 +} + 2 {F}^{-}$

Now ${K}_{\text{sp}} = \frac{\left[B {a}^{2 +}\right] {\left[{F}^{-}\right]}^{2}}{\left[B a {F}_{2} \left(s\right)\right]}$

Now $\left[B a {F}_{2} \left(s\right)\right]$, as a solid, is UNDEFINED, and treated as unity.

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

But we are GIVEN that $\left[B a {F}_{2}\right] = 4.59 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

And thus $\left[B {a}^{2 +}\right] = 4.59 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

And $\left[{F}^{-}\right] = 2 \times 4.59 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

And so ${K}_{\text{sp}} = \left(4.59 \times {10}^{-} 2\right) {\left(2 \times 4.59 \times {10}^{-} 2\right)}^{2}$

$= 4 \times {\left(4.59 \times {10}^{-} 2\right)}^{3} = 3.87 \times {10}^{-} 4$.

This site reports that ${K}_{\text{sp}}$ of $\text{barium fluoride}$ is $1.84 \times {10}^{-} 7$ at $25$ ""^@C. Given the ${K}_{\text{sp}}$ calculated here, which did NOT specify the conditions, would this be at a temperature higher or lower than $298 \cdot K$? Why?