# The solubility of scandium (III) fluoride, ScF_3, in pure water is 2.0 x 10^-5 moles per liter. How do you calculate the value of K_(sp) for scandium (III) fluoride from this data?

Sep 12, 2017

${K}_{\text{sp}} \left(S c {F}_{3}\right) = 4.32 \times {10}^{-} 18$

We interrogate the reaction.....

#### Explanation:

We interrogate the reaction.....

$S c {F}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s S {c}^{3 +} + 3 {F}^{-}$

And as written, we can directly write the solubility expression, ${K}_{\text{sp}}$, where ${K}_{\text{sp}} = \left[S {c}^{3 +}\right] {\left[{F}^{-}\right]}^{3}$

And we are given the solubility of $S c {F}_{3}$ in water, $2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$....

And thus $\left[S {c}^{3 +}\right] = 2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

And $\left[{F}^{-}\right] = 3 \times 2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

Capisce?

And then we input these numbers into the ${K}_{\text{sp}}$ expression.....

${K}_{\text{sp}} = \left(2.0 \times {10}^{-} 5\right) \times {\left(6.0 \times {10}^{-} 5\right)}^{3} = 4.32 \times {10}^{-} 18$

PS I have not got a good text at hand, but the values I read for ${K}_{\text{sp}}$ for scandium fluoride seem to vary widely. I would check the source of these data.