Question

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7}\), or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{<\rho}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\text {s }}=8.1 \times 10^{-12}\), or \(\mathrm{Mn}(\mathrm{OH})_{2}\), \(K_{\text {sp }}=2 \times 10^{-13}\)

Short answer

a. \(S1 = \sqrt[3]{\frac{2.1 \times 10^{-7}}{4}}\), \(S2 = \sqrt[3]{\frac{1.4 \times 10^{-7}}{4}}\), thus \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\) has the smallest molar solubility. b. \(S3 = \sqrt[3]{\frac{8.1 \times 10^{-12}}{4}}\), \(S4 = \sqrt[3]{\frac{2 \times 10^{-13}}{4}}\), thus \(\mathrm{Mn}(\mathrm{OH})_{2}\) has the smallest molar solubility.

Step-by-step solution

(Part a: Write balanced dissolution equations)

(First, we need to write the balanced dissolution equations for both \(\mathrm{FeC}_{2} \mathrm{O}_{4}\) and \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\). For \(\mathrm{FeC}_{2} \mathrm{O}_{4}\): \(\mathrm{FeC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Fe^{2+}}(aq) + 2\mathrm{C_{2}O_{4}^{2-}}(aq)\) For \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\): \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}(s) \rightleftharpoons \mathrm{Cu^{2+}}(aq) + 2\mathrm{IO_{4}^{-}}(aq)\))

(Part a: Derive the expressions for solubility in terms of Ksp)

(Let the molar solubility of \(\mathrm{FeC}_{2} \mathrm{O}_{4}\) be S1 and for \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\) be S2. For \(\mathrm{FeC}_{2} \mathrm{O}_{4}\): \(K_{sp}=[\mathrm{Fe^{2+}}][\mathrm{C_{2}O_{4}^{2-}}]^2 = S1\cdot(2\cdot S1)^2 = S1\cdot(4\cdot S1^2)\) For \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\): \(K_{sp}=[\mathrm{Cu^{2+}}][\mathrm{IO_{4}^{-}}]^2 = S2\cdot(2\cdot S2)^2 = S2\cdot(4\cdot S2^2)\))

(Part a: Calculate molar solubilities for the compounds)

(Now we can solve the expressions for S1 and S2, and then compare their values. For \(\mathrm{FeC}_{2} \mathrm{O}_{4}\): \(S1=\sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2.1 \times 10^{-7}}{4}}\) For \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}\): \(S2=\sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{1.4 \times 10^{-7}}{4}}\)) Part a Answer: Calculate and compare S1 and S2. The compound with the smaller molar solubility will be the one with the smaller S value.

(Part b: Write balanced dissolution equations)

(First, we need to write the balanced dissolution equations for \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Mn}(\mathrm{OH})_{2}\). For \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\): \(\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2\mathrm{Ag^{+}}(aq) + \mathrm{CO_{3}^{2-}}(aq)\) For \(\mathrm{Mn}(\mathrm{OH})_{2}\): \(\mathrm{Mn}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mn^{2+}}(aq) + 2 \mathrm{OH^{-}}(aq)\))

(Part b: Derive the expressions for solubility in terms of Ksp)

(Let the molar solubility of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) be S3 and for \(\mathrm{Mn}(\mathrm{OH})_{2}\) be S4. For \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\): \(K_{sp}=[\mathrm{Ag^{+}}]^2[\mathrm{CO_{3}^{2-}}] = (2\cdot S3)^2\cdot S3 = 4\cdot S3^3\) For \(\mathrm{Mn}(\mathrm{OH})_{2}\): \(K_{sp}=[\mathrm{Mn^{2+}}][\mathrm{OH^{-}}]^2 = S4\cdot(2\cdot S4)^2 = S4\cdot(4\cdot S4^2)\))

(Part b: Calculate molar solubilities for the compounds)

(Now we can solve the expressions for S3 and S4, and then compare their values. For \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\): \(S3=\sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{8.1 \times 10^{-12}}{4}}\) For \(\mathrm{Mn}(\mathrm{OH})_{2}\): \(S4=\sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2 \times 10^{-13}}{4}}\)) Part b Answer: Calculate and compare S3 and S4. The compound with the smaller molar solubility will be the one with the smaller S value.

Key concepts

Solubility Product Constant

The solubility product constant, denoted as Ksp, is a fundamental concept in chemistry that you'll frequently encounter when studying the solubility of slightly soluble salts. Ksp is used to describe the equilibrium between a solid and its ions in solution. At equilibrium, the rate at which the solid dissolves in the solvent is equal to the rate at which the ions recombine to form the solid.

For a generic salt AB that dissociates into A+ and B- ions, the dissolution can be represented as follows:
\[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \]
And the corresponding expression for Ksp would be:
\[ K_{sp} = [A^+][B^-] \]
The square brackets indicate the concentration of the ions in solution. When dealing with more complex salts, such as those with multiple ions, the Ksp expression includes the concentrations of each ion raised to the power of its stoichiometric coefficient. Understanding Ksp allows you to predict whether a precipitate will form under certain conditions and to compare the solubility of different salts, as demonstrated in the exercise.

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time because the forward and reverse reactions occur at the same rate. The concept is not just limited to the dissolution of salts, but applies to all reversible chemical reactions. Equilibrium does not mean the reactants and products are in equal amounts; rather, their ratios don't change with time.

In the context of solubility, equilibrium plays a crucial role in the dissolution of ionic compounds. When a salt is added to a solvent, it begins to dissolve by separating into its constituent ions. As the concentration of these ions increases, some begin to recombine into the original compound. Eventually a dynamic equilibrium is established, where the rate of dissolution equals the rate of precipitation. At this point, the system's Ksp can be determined and used to calculate the molar solubility of the compound. This key principle helps explain why, in the provided exercise, different salts with different Ksp values have different molar solubilities.

Dissolution Equations

Dissolution equations provide a representation of the process by which ionic compounds dissolve and dissociate into their constituent ions in solution. The equations are balanced chemical equations that reflect the stoichiometry of the ionic compound as it separates into cations and anions.

For example, the dissolution of calcium fluoride can be represented as follows:
\[ CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq) \]
The solid compound on the left side of the equation is represented with a (s) for solid, while the ions in solution are indicated with (aq) for aqueous. By defining the solubility (S) of the compound as the molarity of the dissolved ions, the dissolution equation can be used to derive the expression for Ksp. In exercises, such as the ones provided, the dissolution equation is the starting point for calculating the molar solubility of the substance. This is a pivotal step in understanding how to approach problems involving the solubility of ionic compounds and is an excellent example of the practical application of dissolution equations.