Chapter 18.4, Problem 40LC

Interpretation Introduction

Interpretation: The compound $\text{FeS}\left({\text{K}}_{\text{sp}}=8.0\times {10}^{-19}\right)\text{ or }\text{CuS}\left({\text{K}}_{\text{sp}}=8.0\times {10}^{-37}\right)$ having higher solubility should be determined.

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 40LC

  $\text{FeS}$ has a higher solubility.

Explanation of Solution

Given:

  $\begin{array}{l}\text{FeS}\left({\text{K}}_{\text{sp}}=8.0\times {10}^{-19}\right)\\ \text{CuS}\left({\text{K}}_{\text{sp}}=8.0\times {10}^{-37}\right)\end{array}$

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

As the value of the solubility product constant $\text{FeS}$ is higher in comparison to the ${\text{K}}_{\text{sp}}$ of $\text{CuS}$ . Therefore, $\text{FeS}$ has a higher solubility.