Chapter 18, Problem 2STP

Interpretation Introduction

Interpretation: The order of zinc sulfide, zinc carbonate, and silver iodide from most soluble to least stable are to be given.

Concept Introduction: The expression representing the product of the concentrations of the ions each raised to the power (equal to their stoichiometric coefficient) in the dissociation equation is termed as solubility product constant $\left({\text{K}}_{\text{sp}}\right)$ .

Answer to Problem 2STP

The order from most soluble to least soluble is zinc carbonate, silver iodide, and zinc sulfide.

Explanation of Solution

Given:

  ${\text{K}}_{\text{sp}}$of zinc sulfide is $\text{3}\text{.0}\times {\text{10}}^{-23}$

  ${\text{K}}_{\text{sp}}$of zinc carbonate is $\text{1}\text{.0}\times {\text{10}}^{-10}$

  ${\text{K}}_{\text{sp}}$of silver iodide is $\text{8}\text{.3}\times {\text{10}}^{-17}$

The smaller the value of the solubility product constant $\left({\text{K}}_{\text{sp}}\right)$ , the lower is the solubility of the compound.

The value of the solubility product of zinc sulfide is the lowest and the solubility product of zinc carbonate is the highest. Thus, the least soluble is zinc sulfide and zinc carbonate is the most soluble.

Therefore, the order from most soluble to least soluble is zinc carbonate, silver iodide, and zinc sulfide.