# Question #928c8

Feb 20, 2017

Compound $A {X}_{2}$ will have the smallest ${K}_{s} p$ value.

#### Explanation:

The relation between solubility and ${K}_{s p}$ is not a fixed one, but depends on the particular solid. Here's why

If the solid is of form $A X$, it will dissolve, producing one cation ${A}^{+}$ and one anion ${X}^{-}$.

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

and the ${K}_{s p}$ expression is ${K}_{s p} = \left[{A}^{+}\right] \left[{X}^{-}\right]$

If you let $x$ be the solubility, you can show that the ion concentrations will both be equal to $x$, and the ${K}_{s p}$ expression is given by ${x}^{2}$.

On the other hand, both ${A}_{2} X$ and $A {X}_{2}$ dissolve to form three ions, and the ${K}_{s p}$ expression is either

${K}_{s p} = {\left[{A}^{+}\right]}^{2} \left[{X}^{-}\right]$ or ${K}_{s p} = \left[{A}^{+}\right] {\left[{X}^{-}\right]}^{2}$

In either case, using the same assignment of $x$ equal to the solubility, you get to the same expression, the ${K}_{s p}$ expression is equal to $4 {x}^{3}$.

Since all three solubilities are on the order of ${10}^{- 4}$, the ${K}_{s p}$ of AX will be on the order of ${10}^{- 8}$, while the other two will be on the order of ${10}^{- 12}$.

To break the tie, you look at the size of the numbers. $A {X}_{2}$ at $1.20 \times {10}^{- 4}$ is smaller than $1.53 \times {10}^{- 4}$, and so, will generate the smaller ${K}_{s p}$ value.