Problem 50 - Chemistry: The Central Science in SI Units, Global Edition

Question

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, \(4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} .(\mathbf{a})\) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2}\), what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

Step-by-Step Solution

Answer

(a) Salt MZ₂ has the larger solubility product constant, with Ksp₂ = \(128 \times 10^{-12}\) while Ksp₁ = \(16 \times 10^{-8}\) for salt MA. (b) Neither salt has a higher concentration of M²⁺ ions in the saturated solution as both their solubilities are the same. (c) The equilibrium concentration of M²⁺ ions after mixing equal volumes of saturated solutions of MA and MZ₂ is \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\).

Step 1: (a) Determine the solubility product constant

The solubility product constant (Ksp) is calculated using the solubility equilibrium expressions for the given salts. For salt MA: \(MA \rightleftharpoons M^{2+} + A^-\) Ksp₁ = [M²⁺][A⁻] For salt MZ₂: \(MZ_2 \rightleftharpoons M^{2+} + 2Z^-\) Ksp₂ = [M²⁺][Z⁻]² Given the solubility, s = \(4 \times 10^{-4}\, \mathrm{mol\, L^{-1}} \) For salt MA: [M²⁺] = [A⁻] = s = \(4 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Ksp₁ = (\(4 \times 10^{-4}\))^2 = \(16 \times 10^{-8}\) For salt MZ₂: [M²⁺] = s = \(4 \times 10^{-4} \mathrm{mol\, L^{-1}}\) [Z⁻] = 2s = \(8 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Ksp₂ = (\(4 \times 10^{-4}\))(\(8 \times 10^{-4}\))^2 = \(128 \times 10^{-12}\) Comparing the solubility product constants, Ksp₂ > Ksp₁. Therefore, salt MZ₂ has the larger numerical value for the solubility product constant.

Step 2: (b) Determine the higher concentration of M²⁺ in the saturated solution

The solubility of both salts is given as the same, so the concentration of M²⁺ ions in the saturated solution of both salts would also be the same. Thus, neither of them has a higher concentration of M²⁺ ions in the saturated solution.

Step 3: (c) Determine the equilibrium concentration of M²⁺ after mixing the solutions

When mixing equal volumes of saturated solutions of MA and MZ₂, the total volume doubles, and so does the number of moles of each ion. However, the concentration of each ion gets halved. For the mixed solution: [M²⁺'] = \(\frac{1}{2}\) [M²⁺] = \(\frac{1}{2}\) (\(4 \times 10^{-4}\, \mathrm{mol\, L^{-1}}\)) = \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Therefore, the equilibrium concentration of M²⁺ ions after mixing the solutions is \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\).

Key Concepts

Saturated Solution

A saturated solution is defined as a solution in which the maximum amount of solute is dissolved at a given temperature. In other words, no more solute can dissolve in the solvent without changing the temperature or solvent amount. In the case of the two slightly soluble salts, MA and MZ₂, a saturated solution occurs when the amount of each salt that can dissolve has been reached.
When a solution reaches saturation, the rate at which the salt dissolves is equal to the rate at which it precipitates. This establishes a dynamic equilibrium. For MA and MZ₂, each saturates the solution at a solubility of \(4 \times 10^{-4} \ \text{mol/L}\). Beyond this point, any additional salt will remain undissolved.

Solubility Equilibrium

Solubility equilibrium refers to the state where the process of dissolving and the precipitation of a solute occur at equal rates, maintaining the concentration of dissolved ions fixed in a saturated solution. It is represented by equilibrium expressions that vary depending on the dissociation form of the salt.
For example, with the salts MA and MZ₂:

• MA dissociates into \(\text{M}^{2+}\) and \(\text{A}^-\).
• MZ₂ dissociates into \(\text{M}^{2+}\) and \(2\text{Z}^-\).

The solubility product constant \(K_{sp}\) quantifies the equilibrium state for slightly soluble salts. It is calculated as:
MA: \[ K_{sp1} = [\text{M}^{2+}] [\text{A}^-] \]
MZ₂: \[ K_{sp2} = [\text{M}^{2+}] [\text{Z}^-]^2 \]
These expressions help us understand the concentration dynamics of the ions at equilibrium.

Ionic Concentration

Ionic concentration in a solution describes how many ions are present in a given volume of the solution. In a saturated solution, the ionic concentration rises until solubility equilibrium is reached.
Despite MA and MZ₂ having the same solubility of \(4 \times 10^{-4} \ \text{mol/L}\), their dissociation affects ionic concentrations differently. For MA, the concentration of \(\text{M}^{2+}\) is equal to its solubility. However, in MZ₂, the dissociation doubles the concentration of \(\text{Z}^-\) ions.
When the solutions of MA and MZ₂ are mixed, the resulting solution sees a balance due to dilution, halving the ionic concentration. For \(\text{M}^{2+}\), this means its concentration reduces to \(2 \times 10^{-4} \ \text{mol/L}\), assuming equal volumes of the solutions are mixed. Understanding ionic concentration is key to predicting how substances will behave in saturated solutions.