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, \(\mathrm{M}^{2+}\) ?

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.

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.

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 \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\).