If the molar solubility of CaF_2 at 35 C is 1.24 * 10^-3 mol/L, what is Ksp at this temperature?

Jun 9, 2016

${K}_{s p}$ $=$ $\left[C {a}^{2 +}\right] {\left[{F}^{-}\right]}^{2}$ $=$ ??

Explanation:

${K}_{s p}$, $\text{the solubility product}$ derives from the solubility expression:

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

And ${K}_{e q}$ $=$ $\frac{\left[C {a}^{2 +}\right] {\left[{F}^{-}\right]}^{2}}{\left[C a {F}_{2} \left(s\right)\right]}$

However, $\left[C a {F}_{2} \left(s\right)\right]$, the concentration of a solid is a meaningless quantity.

Thus: ${K}_{e q}$ $=$ ${K}_{s p}$ $=$ $\left[C {a}^{2 +}\right] {\left[{F}^{-}\right]}^{2}$

Given this expression, we simply fill in the blanks:

${K}_{s p}$ $=$ $\left(1.24 \times {10}^{-} 3\right) {\left(2 \times 1.24 \times {10}^{-} 3\right)}^{2}$

$=$ $4 \times {\left(1.24 \times {10}^{-} 3\right)}^{3}$

$=$ ???

The given ${K}_{s p}$ is calculated for $35$ ""^@C. At lower temperature, would you expect ${K}_{s p}$ to increase or decrease? Give a reason for your answer.