Question

(a) True or false: “solubility” and “solubility-product constant” are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: MnCO \(_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} .\)

Short answer

(a) False. Solubility and solubility-product constant are not the same number for a given compound. (b) 1. MnCO\(_3\): Ksp = [Mn\(^{2+}\)][CO\(_3^{2-}\)] 2. Hg(OH)\(_2\): Ksp = [Hg\(^{2+}\)][OH\(^{-}\)]\({^2}\) 3. Cu\(_3\)(PO\(_4\))\(_2\): Ksp = [Cu\(^{2+}\)]\({^3}\)[PO\(_4^{3-}\)]\({^2}\)

Step-by-step solution

Answer for Part (a)

For a given compound, solubility and solubility-product constant are not the same. Solubility is a measure of the maximum amount of solute that can dissolve in a solvent at a given temperature and pressure, whereas solubility-product constant (Ksp) is the equilibrium constant for the solubility equilibrium of a slightly soluble compound. The Ksp gives us an idea about the extent of how much of the solid will dissolve in the solution at equilibrium, but it is not the same quantity as solubility. Therefore, the statement is false.

Expressions for Part (b)

To obtain the expressions for the solubility-product constant for each of the given ionic compounds, identify the ions formed when each compound dissolves in water and express the equilibrium in terms of the ion concentrations. 1. For MnCO\(_3\): The dissociation equilibrium can be represented as: MnCO\(_3\)(s) \( \rightleftharpoons \) Mn\(^{2+}\)(aq) + CO\(_3^{2-}\)(aq) The expression for Ksp is then: Ksp = [Mn\(^{2+}\)][CO\(_3^{2-}\)] 2. For Hg(OH)\(_2\): The dissociation equilibrium can be represented as: Hg(OH)\(_2\)(s) \( \rightleftharpoons \) Hg\(^{2+}\)(aq) + 2OH\(^{-}\)(aq) The expression for Ksp is then: Ksp = [Hg\(^{2+}\)][OH\(^{-}\)]\({^2}\) 3. For Cu\(_3\)(PO\(_4\))\(_2\): The dissociation equilibrium can be represented as: Cu\(_3\)(PO\(_4\))\(_2\)(s) \( \rightleftharpoons \) 3Cu\(^{2+}\)(aq) + 2PO\(_4^{3-}\)(aq) The expression for Ksp is then: Ksp = [Cu\(^{2+}\)]\({^3}\)[PO\(_4^{3-}\)]\({^2}\)

Key concepts

Solubility Equilibrium

Solubility equilibrium is a state of balance between a solid and its dissolved ions in a solution. Whenever an ionic compound is placed in water, it begins to dissolve and releases its constituent ions into the solution. However, the process doesn't continue indefinitely. Eventually, enough ions accumulate in the solution that the rate at which they return to form the solid equals the rate at which the solid dissolves. This dynamic balance represents the solubility equilibrium.

• It involves both the dissolved ions and the undissolved solid in equilibrium.
• At this stage, the concentration of ions remains constant over time.

In a solubility equilibrium expression for a compound like MnCO\(_3\), you write the equilibrium as \(\text{MnCO}_3(s) \rightleftharpoons \text{Mn}^{2+}(aq) + \text{CO}_3^{2-}(aq)\). This shows the solid is in equilibrium with its ions in solution. Understanding solubility equilibrium helps predict how changing conditions, like temperature or the addition of more solute, can shift the balance and influence solubility.

Ionic Compounds

Ionic compounds are formed when positive and negative ions join together in a lattice structure due to electrostatic forces. These compounds often dissolve in water, dissociating into ions.

• Each ionic compound has its unique solubility depending on its lattice strength and the ions' charges.
• The process of dissociation involves breaking these ionic bonds, which is why some compounds are more soluble than others.

For example, when MnCO\(_3\) dissolves in water, it separates into Mn\(^{2+}\) and CO\(_3^{2-}\) ions. The dissociation process is essential for calculating the Ksp, which is the solubility product constant. Knowing how ionic compounds behave in solutions allows chemists to predict product formation, reactions, and even water purification processes.

Equilibrium Constant

An equilibrium constant like the solubility product constant (Ksp) describes the position of equilibrium in a chemical reaction. For solubility, the Ksp quantifies the extent to which an ionic compound can dissolve in water.

• The larger the Ksp, the more soluble the compound is.
• Ksp values are determined experimentally and are specific to each compound under specific conditions.

To construct the Ksp for a compound such as Cu\(_3\)(PO\(_4\))\(_2\), one writes the dissociation equation and uses the concentrations of the ions in the equilibrium expression: \(\text{Ksp} = [\text{Cu}^{2+}]^3[\text{PO}_4^{3-}]^2\). Calculating the Ksp involves using concentrations at equilibrium, providing a powerful tool for predicting solubility and understanding reaction dynamics.