Question

The cobalt(II) ion concentration in a saturated solution of cobalt(II) sulfide, \(\operatorname{CoS},\) is \(7.7 \times 10^{-11} \mathrm{M}\). Calculate the value for the equilibrium constant.

Short answer

The equilibrium constant, \( K_{sp} \), for cobalt(II) sulfide is \( 5.93 \times 10^{-21} \).

Step-by-step solution

Write the Dissolution Equation

The dissolution of cobalt(II) sulfide, \( \text{CoS} \), in water is represented by the equilibrium reaction: \[ \text{CoS(s)} \rightleftharpoons \text{Co}^{2+}\text{(aq)} + \text{S}^{2-}\text{(aq)} \] In this equation, \( \text{CoS} \) is a solid, and its concentration is not included in the equilibrium expression.

Write the Expression for the Solubility Product Constant (Ksp)

The solubility product constant, \( K_{sp} \), for the dissolution of \( \text{CoS} \) can be expressed as: \[ K_{sp} = [\text{Co}^{2+}][\text{S}^{2-}] \] where \([\text{Co}^{2+}]\) and \([\text{S}^{2-}]\) are the molar concentrations of the cobalt(II) and sulfide ions at equilibrium, respectively.

Use Ion Concentrations to Determine Ksp

Given that the concentration of \( \text{Co}^{2+} \) is \(7.7 \times 10^{-11} \ \mathrm{M}\), and assuming \([\text{Co}^{2+}] = [\text{S}^{2-}]\), since each mole of \( \text{CoS} \) gives one mole of \( \text{Co}^{2+} \) and one mole of \( \text{S}^{2-} \), the concentration of \( \text{S}^{2-} \) is also \(7.7 \times 10^{-11} \ \mathrm{M}\).

Calculate Ksp

Substitute the values of ion concentrations into the expression for \( K_{sp} \): \[ K_{sp} = (7.7 \times 10^{-11})(7.7 \times 10^{-11}) \] \[ K_{sp} = 5.93 \times 10^{-21} \] Therefore, the equilibrium constant, \( K_{sp} \), is \( 5.93 \times 10^{-21} \).

Key concepts

Cobalt(II) Sulfide

Cobalt(II) sulfide, represented by the chemical formula \( \text{CoS} \), is an ionic compound consisting of cobalt and sulfide ions. Cobalt belongs to the transition metal category and possesses unique properties compared to more common metals, like those we see in iron or copper. Being a compound composed of a metal and a non-metal, cobalt(II) sulfide forms a solid crystalline lattice at room temperature.

In aqueous solutions, this solid is not very soluble, meaning only a tiny fraction of it dissolves into its constituent ions: the cobalt ions \( \text{Co}^{2+} \) and the sulfide ions \( \text{S}^{2-} \). This limited solubility makes it important to understand how cobalt(II) sulfide behaves as it influences the concentration of ions in a solution. As only a small amount dissolves, this makes cobalt(II) sulfide useful in industries needing precise low-concentration metal ion solutions, for example in electroplating or as catalysts.

Solubility Product Constant

The solubility product constant, commonly abbreviated as \( K_{sp} \), plays a crucial role in predicting the solubility of ionic compounds in water. Specifically, \( K_{sp} \) is a measure of the extent at which a compound dissociates into its ions when it dissolves in water. It gives insight into the concentration of ions that can be present in a saturated solution for a given ionic compound. Put simply, the higher the \( K_{sp} \), the more soluble a substance is.

For cobalt(II) sulfide, the expression for \( K_{sp} \) is defined as the product of the molar concentrations of the dissolved ions, specifically \([\text{Co}^{2+}][\text{S}^{2-}]\). Since cobalt(II) sulfide dissolves poorly, its \( K_{sp} \) value is quite low, which reflects its low solubility in water. By calculating \( K_{sp} \), students can predict how much cobalt(II) sulfide will dissolve, which is essential for applications in chemistry and chemical engineering.

Dissolution Equation

The dissolution equation is a representation of what happens when a solid compound, like cobalt(II) sulfide, comes into contact with water. In the dissolution equation \( \text{CoS(s)} \rightleftharpoons \text{Co}^{2+}\text{(aq)} + \text{S}^{2-}\text{(aq)} \), the solid \( \text{CoS} \) dissociates into its respective ions. In this reaction, the solid state of cobalt(II) sulfide is on the left of the arrows, while the aqueous (dissolved in water) phases of the cobalt and sulfide ions are on the right.

Understanding the dissolution equation helps illustrate how equilibrium is achieved in a saturated solution. At equilibrium, the rate at which \( \text{CoS} \) dissolves is equal to the rate at which the dissolved ions recombine to form \( \text{CoS} \). This balance ensures that the concentrations of \( \text{Co}^{2+} \) and \( \text{S}^{2-} \) ions remain constant over time, providing a predictable measure for calculation using the solubility product constant.