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

Question

Olivia Anton · Accepted Answer

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

We scrutinize the balance:

#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?

Wyatt Atkins · Answer

undefined The solubility product constant (Ksp) for barium fluoride (BaF2) is calculated using the formula:

$$ K_{sp} = [Ba^{2+}][F^-]^2 $$

Given the solubility of BaF2 as 4.59 × 10^(-2) mol/L, the concentration of Ba^2+ ions and F^- ions in the saturated solution is equal to the solubility. Therefore:

$$ [Ba^{2+}] = [F^-] = 4.59 \times 10^{-2} \, \text{mol/L} $$

Substituting these values into the Ksp expression:

$$ K_{sp} = (4.59 \times 10^{-2}) \times (4.59 \times 10^{-2})^2 $$

$$ K_{sp} = (4.59 \times 10^{-2}) \times (2.10981 \times 10^{-3}) $$

$$ K_{sp} = 9.6824 \times 10^{-5} \, \text{mol}^3/\text{L}^3 $$

Therefore, the solubility product constant (Ksp) for barium fluoride (BaF2) is $ 9.6824 \times 10^{-5} \, \text{mol}^3/\text{L}^3 $.