Problem 48 - Chemistry: The Central Science

Question

(a) Explain the difference between solubility and solubility-product constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2}\), and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Step-by-Step Solution

Answer

(a) Solubility represents the maximum amount of a solute that can dissolve in a solvent (given in units like g/L or mol/L), while the solubility-product constant (\(K_{sp}\)) is an equilibrium constant representing the extent of an ionic compound's dissolution in water (unitless). (b) For the given ionic compounds, the expressions for the solubility-product constants are: \(K_{sp}(\mathrm{MnCO}_3) = [\mathrm{Mn^{2+}}] [\mathrm{CO_3^{2-}}]\) \(K_{sp}(\mathrm{Hg(OH)_2}) = [\mathrm{Hg^{2+}}] [\mathrm{OH^-}]^2\) \(K_{sp}(\mathrm{Cu_{3}(PO_4)_2}) = [\mathrm{Cu^{2+}}]^3 [\mathrm{PO_4^{3-}}]^2\)

Step 1: Definition of Solubility

Solubility is the maximum amount of a solute that can dissolve in a certain amount of solvent at a specific temperature and pressure. It is generally given in units such as grams of solute per liter of solvent (g/L) or moles of solute per liter of solvent (mol/L).

Step 2: Definition of Solubility-Product Constant

The solubility-product constant, denoted by \(K_{sp}\), is an equilibrium constant that represents the extent to which an ionic compound dissolves in water to form a saturated solution. It is equal to the product of the concentrations of the ions formed, each raised to the power of their stoichiometric coefficients, at equilibrium.

Step 3: Difference Between Solubility and Solubility-Product Constant

The main difference between solubility and solubility-product constant is that solubility represents the maximum amount of solute that can dissolve in a solvent, while the solubility-product constant represents the equilibrium constant of the dissolution process. Also, solubility has specific units (e.g., g/L or mol/L), while the solubility-product constant is unitless.

Step 4: Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{MnCO}_3\)

The dissolution equation for \(\mathrm{MnCO}_3\) is: \(\mathrm{MnCO}_3 (s) \rightleftharpoons \mathrm{Mn^{2+}} (aq) + \mathrm{CO_3^{2-}} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{MnCO}_{3}\) is: \(K_{sp} = [\mathrm{Mn^{2+}}] [\mathrm{CO_3^{2-}}]\)

Step 5: Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{Hg}(\mathrm{OH})_{2}\)

The dissolution equation for \(\mathrm{Hg}(\mathrm{OH})_{2}\) is: \(\mathrm{Hg}(\mathrm{OH})_2 (s) \rightleftharpoons \mathrm{Hg^{2+}} (aq) + 2 \mathrm{OH^-} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{Hg}(\mathrm{OH})_{2}\) is: \(K_{sp} = [\mathrm{Hg^{2+}}] [\mathrm{OH^-}]^2\)

Step 6: Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

The dissolution equation for \(\mathrm{Cu}_{3}\left(\mathrm{PO}_4\right)_2\) is: \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} (s) \rightleftharpoons 3 \mathrm{Cu^{2+}} (aq) + 2 \mathrm{PO_4^{3-}} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \(K_{sp} = [\mathrm{Cu^{2+}}]^3 [\mathrm{PO_4^{3-}}]^2\)

Key Concepts

Solubility

Imagine pouring sugar into a cup of tea; there is a limit to how much will dissolve, and that limit is what we call solubility. It represents how much solute can be dissolved in a solvent to form a homogeneous mixture at a specific temperature and pressure. Usually, solubility is expressed in terms like grams per liter (g/L) or moles per liter (mol/L). This concept is significant because it dictates the extent to which a substance can be dissolved in a solvent before reaching a saturated solution.

For instance, when we stir a spoonful of sugar in water, we're witnessing an increase in the solubility of sugar until no more can dissolve, reaching what is known as saturation. Beyond this point, any additional sugar will just settle at the bottom. With temperature changes, solubility can also vary; often, heat will increase solubility for many substances allowing more to dissolve.

Equilibrium Constant

To delve into the equilibrium constant, we must understand that chemical reactions can reach a state where the forward and reverse reactions occur at the same rate. When that happens, we've reached equilibrium. The equilibrium constant, abbreviated as 'K', gives us a peek into the balance between reactants and products in this steady state.

For the dissolution of ionic compounds in water, we focus on a special type of equilibrium constant known as the solubility-product constant, symbolized by (K_{sp}). It has no units and considers only the ions that are formed from the ionic compound in a saturated solution. Higher (K_{sp}) values signify greater solubility, and hence, a greater degree of dissociation into ions. This key piece of information is crucial in predicting the formation of a precipitate, understanding solution behavior, and manipulating separation processes like selective precipitation.

Dissolution Equations

Dissolution equations are the backbone of understanding how compounds separate into ions in a solution. Simply put, these equations provide a visual formula of the disintegration process of compounds when they dissolve in a solvent. Let's break down a complex process into an easy concept. Think of dropping a sugar cube into a cup of warm water. As it dissolves, the sugar molecules separate and spread throughout the water until evenly distributed.

In the case of ionic compounds, such as salt (NaCl), the dissolution process involves separation into ions (Na+ and Cl-). These ions will individually dissolve in water. The dissolution equation depicts this process, and when combined with the concept of the equilibrium constant, or (K_{sp}), we get a quantitative measure of this solubility behavior. Consequently, these dissolution equations and their constants are integral in predicting and explaining the outcomes of many reactions in aqueous solutions.