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

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: \(K_{sp, MZ_{2}} = [\mathrm{M}^{2+}][\mathrm{Z}^{-}]^{2}\)

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 \(\mathrm{M}^{2+}.\)

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 \(4 \times 10^{-4} \mathrm{mol/L}\).