# Question #0e38d

May 15, 2015

The solubility product constant for this salt will be $1.22 \cdot {10}^{- 3}$.

When dissolved in water, your salt will dissociate into cations and anions according to the following equilibrium

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

Since you're dealing with a 1.00-L solution, the molar solubility of your salt will be

$C = \frac{n}{V} = \text{0.0820 moles"/"1.00 L" = "0.0820 mol/L}$

Take a look at the mole ratio that exists between the species involved in the equilibrium. You have 1 mole of $A {B}_{3}$ dissociating to produce 1 mole of ${A}^{3 +}$ and $\textcolor{red}{3}$ moles of ${B}^{-}$.

You can use an ICE table to determine the value of the ${K}_{s p}$.

$\text{ } A {B}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s {A}_{\left(a q\right)}^{3 +} + {B}_{\left(a q\right)}^{-}$
I........$-$.............0..............0
C......$-$............(+x)...........(+$\textcolor{red}{3}$x)
E.......$-$.............x...............3x

By definition, ${K}_{s p}$ will be

${k}_{s p} = \left[{A}^{3 +}\right] \cdot {\left[{B}^{-}\right]}^{\textcolor{red}{3}} = x \cdot {\left(3 x\right)}^{3} = 27 \cdot {x}^{4}$

But $x$ is actually the molar solubility of the salt, 0.0820 mol/L, which means that the value of the solubility product constant will be

${K}_{s p} = 27 \cdot {\left(0.0820\right)}^{4} = \textcolor{g r e e n}{1.22 \cdot {10}^{- 3}}$