# A salt, AB_3, expresses an aqueous solubility of 0.0750*mol*L^-1 with respect to the given solubility? What is K_"sp" for this solute?

Nov 22, 2016

${K}_{s p} = \left[{A}^{3 +}\right] {\left[{B}^{-}\right]}^{3} = 27 \times {0.0750}^{4}$

#### Explanation:

We examine the following equilibrium:

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

Since the solid cannot express a concentration, we write the following expression to represent the $\text{solubility product:}$

${K}_{s p} = \left[{A}^{3 +}\right] {\left[{B}^{-}\right]}^{3}$

As is usual in these problems, we raise each concentration to the power of the stoichiometric coefficient in the solubility expression, and thus we have the product ${\left[{A}^{3 +}\right]}^{1} \times {\left[{B}^{-}\right]}^{3}$.

But we are given that $\left[A {B}_{3} \left(a q\right)\right] = 0.0750 \cdot m o l \cdot {L}^{-} 1 ,$ and thus we are in a position to evaluate ${K}_{\text{sp}}$ directly.

And thus $\left[A\right] = 0.0750 \cdot m o l \cdot {L}^{-} 1 ,$ and $\left[{B}^{-}\right] = 3 \times 0.0750 \cdot m o l \cdot {L}^{-} 1$.

${K}_{s p} = \left[0.0750\right] {\left[3 \times 0.0750\right]}^{3} = 27 \times {0.0750}^{4}$

Are you happy with this?