Question #928c8

1 Answer
Feb 20, 2017

Compound AX_2AX2 will have the smallest K_spKsp value.

Explanation:

The relation between solubility and K_(sp)Ksp is not a fixed one, but depends on the particular solid. Here's why

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

AX(s) rightleftharpoons A^+ + X^-AX(s)A++X

and the K_(sp)Ksp expression is K_(sp)=[A^+][X^-]Ksp=[A+][X]

If you let xx be the solubility, you can show that the ion concentrations will both be equal to xx, and the K_(sp)Ksp expression is given by x^2x2.

On the other hand, both A_2XA2X and AX_2AX2 dissolve to form three ions, and the K_(sp)Ksp expression is either

K_(sp)=[A^+]^2[X^-]Ksp=[A+]2[X] or K_(sp)=[A^+][X^-]^2Ksp=[A+][X]2

In either case, using the same assignment of xx equal to the solubility, you get to the same expression, the K_(sp)Ksp expression is equal to 4x^34x3.

Since all three solubilities are on the order of 10^(-4)104, the K_(sp)Ksp of AX will be on the order of 10^(-8)108, while the other two will be on the order of 10^(-12)1012.

To break the tie, you look at the size of the numbers. AX_2AX2 at 1.20xx10^(-4)1.20×104 is smaller than 1.53xx10^(-4)1.53×104, and so, will generate the smaller K_(sp)Ksp value.