Question

Solubility products predict that cation ${\mathbf{AX}}^{\mathbf{3}\mathbf{+}}$can be 99.999% separated from cation ${\mathbf{B}}^{\mathbf{2}\mathbf{+}}$by precipitation with anion X . When the separation is tried, we find 0.2% contamination of ${\mathbf{AX}}_{\mathbf{3}}\left(s\right)$with ${\mathbf{B}}^{\mathbf{2}\mathbf{+}}$. Explain what might be happening.

Short answer

${\mathrm{BX}}_2$is coprecipitated with ${\mathrm{AX}}_3$. This indicates some amount of ${\mathrm{BX}}_2$is surrounded in the ${\mathrm{AX}}_3$during the course of precipitation of ${\mathrm{AX}}_3$. So, we have found the contamination.

Step-by-step solution

Definition of a solubility product

For a sparingly soluble salt, the solubility product is the mathematical product of the concentration of it dissolved constituent ions raised up to the power of their stoichiometric coefficient.

Definition of a solubility product

For a sparingly soluble salt, the solubility product is the mathematical product of the concentration of it dissolved constituent ions raised up to the power of their stoichiometric coefficient.

$$\begin{gathered} M_y{X}_{(}z\left(s\right))\rightleftharpoons y{M}^z+\left(aq\right)+z{X}^{y-}\left(aq\right) \\ {K}_{sp}={\left[M^{z+}\right]}^y{\left[X^{y-}\right]}^z \end{gathered}$$

Explanation for impurity

In a mixture of two salts with the same anion the solubility product helps in predicting the precipitation of a particular salt. The theoretical yield for the precipitation of ${\mathrm{AX}}_3$is 99.999% . The reaction quotient for ${\mathrm{BX}}_2$must have exceeded the solubility product for ${\mathrm{BX}}_2$. Therefore some amount of has coprecipitated with ${\mathrm{AX}}_3$. So, we have found the contamination.