Problem 40 - Chemistry

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 \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2},\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

Step-by-Step Solution

Answer

a. Molar solubility of Cu(IO4)2 < molar solubility of FeC2O4. b. Molar solubility of Ag2CO3 < molar solubility of Mn(OH)2.

Step 1: a. FeC2O4 and Cu(IO4)2

First, let's write down the dissolution equations and expressions for molar solubility for each solid: FeC2O4 ⟶ Fe^2+ + 2C2O4^2- , molar solubility (FeC2O4) = x Equilibrium constant \(K_{sp} = [Fe^{2+}][C2O4^{2-}]^2 = x * (2x)^2 = (4x^3)\) Cu(IO4)2 ⟶ Cu^2+ + 2IO4^-, molar solubility (Cu(IO4)2) = y Equilibrium constant \(K_{sp} = [Cu^{2+}][IO4^-]^2 = y * (2y)^2 = (4y^3)\) Now we will compare: (4x^3) = \(2.1 \times 10^{-7}\) x = \(\surd[2.1 \times 10^{-7}/4]^{1/3}\) (4y^3) = \(1.4 \times 10^{-7}\) y = \(\surd[1.4 \times 10^{-7}/4]^{1/3}\) As x > y, then molar solubility of Cu(IO4)2 (y) is smaller.

Step 2: b. Ag2CO3 and Mn(OH)2

Again, let's write down the dissolution equations and expressions for molar solubility for each solid: Ag2CO3 ⟶ 2Ag^+ + CO3^2-, molar solubility (Ag2CO3) = x Equilibrium constant \(K_{sp} = [Ag^+]^2[CO3^{2-}] = (2x)^2 * x = (4x^3)\) Mn(OH)2 ⟶ Mn^2+ + 2OH^-, molar solubility (Mn(OH)2) = y Equilibrium constant \(K_{sp} = [Mn^{2+}][OH^-]^2 = y * (2y)^2 = (4y^3)\) Now we will compare: (4x^3) = \(8.1 \times 10^{-12}\) x = \(\surd[8.1 \times 10^{-12}/4]^{1/3}\) (4y^3) = \(2 \times 10^{-13}\) y = \(\surd[2 \times 10^{-13}/4]^{1/3}\) As x < y, then molar solubility of Ag2CO3 (x) is smaller.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, often denoted as \(K_{sp}\), is a crucial concept when it comes to understanding the solubility of ionic compounds in water. It represents the maximum product of the concentrations of the ions produced by the dissociation of a compound at equilibrium. This constant is unique to each compound and depends on temperature.

• For example, if a compound dissolves to give ions \(A^+\) and \(B^-\), then \(K_{sp} = [A^+][B^-]\).
• A higher \(K_{sp}\) value typically indicates a more soluble compound.
• Conversely, a lower \(K_{sp}\) suggests a compound is less soluble in water.

Understanding \(K_{sp}\) helps in predicting whether a precipitate will form in a solution when mixed with another. This concept is frequently used in comparing the solubility of different solutes.

Dissolution Equations

Dissolution equations are straightforward representations of how a solid ionic compound disassociates into its component ions when it is dissolved in water. These equations are key to deriving expressions for molar solubility and equilibrium constant of the compound. In the exercise, we looked at substances like FeC2O4 and Cu(IO4)2. For each of these, we first write their dissolution equation:

• FeC2O4 dissolves as \(\mathrm{FeC}_{2}\mathrm{O}_{4} \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\).
• Cu(IO4)2 dissolves as \(\mathrm{Cu}(\mathrm{IO}_{4})_{2} \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{IO}_{4}^{-}\).

The stoichiometry in these equations shows the ratio in which the solute dissociates into ions. This directly affects the calculation of the solubility product, involving components like \(x\) or \(y\), which represent molar solubility of each compound.

Equilibrium Calculations

Equilibrium calculations are crucial in determining the molar solubility of a compound in a given solution. The goal is to calculate the concentration of ions at which the solution is saturated, meaning no more solute can be dissolved without forming a precipitate. Given the \(K_{sp}\), we can set up equilibrium expressions for the dissociation of compounds. For example:

• For FeC2O4, with equilibrium expression \([\mathrm{Fe}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\), we get \(K_{sp} = 4x^3\).
• Similarly, for Mn(OH)2, with \([\mathrm{Mn}^{2+}][\mathrm{OH}^{-}]^2\), we have \(K_{sp} = 4y^3\).

Through equilibrium calculations, we solve for \(x\) and \(y\) to find the molar solubility. These calculations involve basic algebraic manipulation and the use of cube roots. They highlight how the number of ions produced from dissolution affects the final solubility determination.