Question

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2},\) is the same, \(4 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) . (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 \(\mathrm{M}^{2+} ?\) (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, \(\mathrm{M}^{2+}\) ?

Short answer

(a) The salt with the larger numerical value for the solubility product constant is \(\mathrm{MZ}_{2}\). (b) Both salts have the same concentration of \(\mathrm{M}^{2+}\), which is \(4 \times 10^{-4} \mathrm{mol/L}\). (c) The equilibrium concentration of the cation, \(\mathrm{M}^{2+}\), after mixing equal volumes of the saturated solutions is \(4 \times 10^{-4} \mathrm{mol/L}\).

Step-by-step solution

Write the dissolution equations and Ksp expression for both salts

For salt \(\mathrm{MA}\): Dissolution equation: \(\mathrm{MA} \rightleftharpoons \mathrm{M}^{2+} + \mathrm{A}^{-}\) Ksp expression: \(K_{sp, MA} = [\mathrm{M}^{2+}][\mathrm{A}^{-}]\) For salt \(\mathrm{MZ}_{2}\): Dissolution equation: \(\mathrm{MZ}_{2} \rightleftharpoons \mathrm{M}^{2+} + 2 \mathrm{Z}^{-}\) Ksp expression: \(K_{sp, MZ_{2}} = [\mathrm{M}^{2+}][\mathrm{Z}^{-}]^{2}\)

Calculate the Ksp values for both salts

We are given that the solubility of both salts is the same, \(4 \times 10^{-4} \mathrm{mol/L}\). Let's use this information to find \(K_{sp}\) values for both salts. For salt \(\mathrm{MA}\): Solubility: \([\mathrm{M}^{2+}] = [\mathrm{A}^{-}] = 4 \times 10^{-4} \mathrm{mol/L}\) \(K_{sp, MA} = (4 \times 10^{-4})(4 \times 10^{-4}) = 16 \times 10^{-8}\) For salt \(\mathrm{MZ}_{2}\): Solubility: \([\mathrm{M}^{2+}] = 4 \times 10^{-4} \mathrm{mol/L}\), \([\mathrm{Z}^{-}] = 2(4 \times 10^{-4}) \mathrm{mol/L} = 8 \times 10^{-4} \mathrm{mol/L}\) \(K_{sp, MZ_{2}} = (4 \times 10^{-4})(8 \times 10^{-4})^{2} = 4 \times 10^{-4} \times 64 \times 10^{-8} = 256 \times 10^{-12}\) (a) Since \(K_{sp, MZ_{2}} > K_{sp, MA}\), the salt with the larger numerical value for the solubility product constant is \(\mathrm{MZ}_{2}\). (b) In a saturated solution of each salt, both \(\mathrm{MA}\) and \(\mathrm{MZ}_{2}\) have the same concentration of M\(^{2+}\), which is \(4 \times 10^{-4} \mathrm{mol/L}\). Therefore, both have the same concentration of \(\mathrm{M}^{2+}.\)

Determine the equilibrium concentration of M\(^{2+}\) after mixing equal volumes of saturated solutions of both salts

Let \(V\) be the volume of each saturated solution. When mixed, the volume doubles, but since the solubility of both salts is the same, no precipitation will occur. The concentration of M\(^{2+}\) before mixing in both solutions is \(4 \times 10^{-4} \mathrm{mol/L}\). Upon mixing, the moles of M\(^{2+}\) are preserved, and hence, in the combined 2\(V\) volume, the equilibrium concentration of M\(^{2+}\) remains unchanged at \(4 \times 10^{-4} \mathrm{mol/L}\). (c) The equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) after mixing equal volumes of the saturated solutions is \(4 \times 10^{-4} \mathrm{mol/L}\).

Key concepts

Dissolution Equations

When dealing with slightly soluble salts like \( \text{MA} \) and \( \text{MZ}_2 \), understanding their dissolution equations is crucial. These equations show how the salts dissociate into ions when they dissolve in water.

For the salt \( \text{MA} \), the dissolution equation is:
\[ \text{MA} \rightleftharpoons \text{M}^{2+} + \text{A}^- \] This means that one unit of \( \text{MA} \) dissolves to form one \( \text{M}^{2+} \) ion and one \( \text{A}^- \) ion. The dissolution reactions must be understood because they help determine the stoichiometry of ions in a solution, important for calculating further chemical constants like the solubility product.

For \( \text{MZ}_2 \), the equation is different due to its composition:
\[ \text{MZ}_2 \rightleftharpoons \text{M}^{2+} + 2\text{Z}^- \] Here, \( \text{MZ}_2 \) dissolves to provide one \( \text{M}^{2+} \) ion but two \( \text{Z}^- \) ions. Because each formula unit of \( \text{MZ}_2 \) produces two \( \text{Z}^- \) ions, this affects both the concentration balance in the solution and the calculation of its solubility product, \( K_{sp} \).

Saturated Solution

A saturated solution is one where the maximum amount of solute has been dissolved in a solvent, and any additional solute will not dissolve. In this context, when \( \text{MA} \) or \( \text{MZ}_2 \) is fully dissolved, the solution reaches saturation.

Saturated solutions are critical in understanding the equilibrium between solid and dissolved ions. In saturation, the rates of dissolution and precipitation of the ions become equal, allowing us to define the solubility product constant \( K_{sp} \).

For both salts in the exercise, they reach saturation at a concentration of \(4 \times 10^{-4} \text{ mol/L} \). This indicates that any additional ions added would not dissolve until equilibrium shifts due to changes in concentration, temperature, or pressure.

The constant concentration in a saturated solution is key to predicting how the solution will behave under different conditions, such as when mixed with another solution.

Cation Concentration

Cation concentration refers to the amount of positively charged ions present in solution. In our exercise, the focus is on the concentration of \( \text{M}^{2+} \). The dissolution equations help determine the initial concentration of these ions in the solution.

When both \( \text{MA} \) and \( \text{MZ}_2 \) dissolve, they release \( \text{M}^{2+} \) ions. Despite \( \text{MZ}_2 \) having a different formula and dissolution pattern, its \( \text{M}^{2+} \) concentration remains the same as \( \text{MA} \) at \( 4 \times 10^{-4} \text{ mol/L} \) in a saturated solution.

It's important to realize that although the cation concentrations are equal, the solubility product values differ due to the differing stoichiometry.

When mixing solutions, the cation concentration in the resulting mixture remains crucial for predicting reactivity and precipitation. Maintaining the understanding that \( \text{M}^{2+} \) ions do not change in concentration although the total volume has doubled is critical when dealing with equilibrium conditions.