# In a certain mystery liquid, the compounds Al_2S_3, K_3N, Na_2S, and RbBr each have a value of Ksp = 1.0 xx 10^-10. Which of these solids should be the least soluble in the mystery liquid? (Choices in answer).

Aug 4, 2015

The answer is indeed A. rubidium bromide

#### Explanation:

The options are:

A. $R b B r$
B. $N {a}_{2} S$
C. ${K}_{3} N$
D. $A {l}_{2} {S}_{3}$
E. All four solids should be equally soluble in
the liquid since their ${K}_{s p}$’s are all equal

The idea behind this problem is very simple - you have to use the definition of the solubility product constant, ${K}_{s p}$, to determine which compound has the greatest molar solubility.

An ionic compound's solubility is measured by its degree of dissociation. That means that a more soluble ionic compound will produce greater concentrations of cations and anions in solution.

${K}_{s p}$ simply tells you where the dissociation equilibrium lies, or, more precisely, how far to the left it lies.

Take a look at the dissociation equilibria for your compounds

$A {l}_{2} {S}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s 2 A {l}_{\left(a q\right)}^{3 +} + 3 {S}_{\left(a q\right)}^{2 -}$, ${K}_{s p 1} = {\left[A {l}^{3 +}\right]}^{2} \cdot {\left[{S}^{2 -}\right]}^{3}$

${K}_{3} {N}_{\left(s\right)} r i g h t \le f t h a r p \infty n s 3 {K}_{\left(a q\right)}^{+} + {N}_{\left(a q\right)}^{3 -}$, ${K}_{s p 2} = {\left[{K}^{+}\right]}^{3} \cdot \left[{N}^{3 -}\right]$

$N {a}_{2} {S}_{\left(s\right)} r i g h t \le f t h a r p \infty n s 2 N {a}_{\left(a q\right)}^{+} + {S}_{\left(a q\right)}^{2 -}$, ${K}_{s p 3} = {\left[N {a}^{+}\right]}^{2} \cdot \left[{S}^{2 -}\right]$

$R b B {r}_{\left(s\right)} r i g h t \le f t h a r p \infty n s R {b}_{\left(a q\right)}^{+} + B {r}_{\left(a q\right)}^{-}$, ${K}_{s p 4} = \left[R b\right] \cdot \left[B r\right]$

If the ${K}_{s p}$ values are equal for all these reactions, then you could predict that the compound that produces that smallest number of cations and anions per liter will be the least soluble.

Even without doing any calculations, you can predict that $A {l}_{2} {S}_{3}$ will be the most soluble and $R b B r$ will be the least soluble.

You can determine the molar solubility of each of those compounds by replacing the concentrations of the cations and anions with $x$.

This will get you

${K}_{s p 1} = {\left(3 x\right)}^{2} \cdot {\left(2 x\right)}^{3} = 72 {x}^{5}$

${K}_{s p 2} = {\left(3 x\right)}^{3} \cdot x = 27 {x}^{4}$

${K}_{s p 3} = {\left(2 x\right)}^{2} \cdot x = 4 {x}^{3}$

${K}_{s p 4} = x \cdot x = {x}^{2}$

Now it becomes obvious that rubidium bromide will have the lowest molar solubility, i.e. the smallest $x$ value.

This means that one mole of $R b B r$ will produce less $R {b}^{+}$ and $B {r}^{-}$ ions per liter than 1 mole of $N {a}_{2} S$, which in turn will produce less ions in solution than one mole of ${K}_{3} N$, which in turn will produce less ions in solution than $A {l}_{2} {S}_{3}$.