The solubility of scandium (III) fluoride, $S c F_{3}$ $ScF_3$ , in pure water is $2.0 x 10^{- 5}$ $2.0 x 10^-5$ moles per liter. How do you calculate the value of $K_{s p}$ $K_(sp)$ for scandium (III) fluoride from this data?

Question

The solubility of scandium (III) fluoride, $S c F_{3}$ $ScF_3$ , in pure water is $2.0 x 10^{- 5}$ $2.0 x 10^-5$ moles per liter. How do you calculate the value of $K_{s p}$ $K_(sp)$ for scandium (III) fluoride from this data?

1 Answer

Impact of this question

7850 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-12T17:15:54

$K_{sp} (S c F_{3}) = 4.32 \times 10^{- 18}$ $K_"sp"(ScF_3)=4.32xx10^-18$

We interrogate the reaction.....

Explanation:

We interrogate the reaction.....

$S c F_{3} (s) ⇌ S c^{3 +} + 3 F^{-}$ $ScF_3(s) rightleftharpoons Sc^(3+) + 3F^-$

And as written, we can directly write the solubility expression, $K_{sp}$ $K_"sp"$ , where $K_{sp} = [S c^{3 +}] {[F^{-}]}^{3}$ $K_"sp"=[Sc^(3+)][F^-]^3$

And we are given the solubility of $S c F_{3}$ $ScF_3$ in water, $2.0 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $2.0xx10^-5*mol*L^-1$ ....

And thus $[S c^{3 +}] = 2.0 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[Sc^(3+)]=2.0xx10^-5*mol*L^-1$

And $[F^{-}] = 3 \times 2.0 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[F^-]=3xx2.0xx10^-5*mol*L^-1$

Capisce?

And then we input these numbers into the $K_{sp}$ $K_"sp"$ expression.....

$K_{sp} = (2.0 \times 10^{- 5}) \times {(6.0 \times 10^{- 5})}^{3} = 4.32 \times 10^{- 18}$ $K_"sp"=(2.0xx10^-5)xx(6.0xx10^-5)^3=4.32xx10^-18$

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

Science

Math

Humanities

... and beyond

The solubility of scandium (III) fluoride, $S c F_{3}$ $ScF_3$ , in pure water is $2.0 x 10^{- 5}$ $2.0 x 10^-5$ moles per liter. How do you calculate the value of $K_{s p}$ $K_(sp)$ for scandium (III) fluoride from this data?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

The solubility of scandium (III) fluoride, ScF3ScF_3, in pure water is 2.0x10−52.0 x 10^-5 moles per liter. How do you calculate the value of KspK_(sp) for scandium (III) fluoride from this data?

1 Answer

Explanation:

Related questions

Impact of this question

The solubility of scandium (III) fluoride, $S c F_{3}$ $ScF_3$ , in pure water is $2.0 x 10^{- 5}$ $2.0 x 10^-5$ moles per liter. How do you calculate the value of $K_{s p}$ $K_(sp)$ for scandium (III) fluoride from this data?