Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and $\mathrm{Cd}^{2+},(\mathbf{b}) \mathrm{Cu}^{2+}\( and \)\mathrm{Mg}^{2+},(\mathbf{c}) \mathrm{Pb}^{2+}$ and \(\mathrm{Al}^{3+},(\mathbf{d}) \mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+} .\)

To separate Na+ and Cd2+ ions from a solution mixture, we can take advantage of the difference in solubility of their respective sulfide salts. Step 1: Add a source of sulfide ions, such as sodium sulfide (Na2S), to the solution containing the Na+ and Cd2+ ions. This will cause the formation of insoluble cadmium sulfide (CdS) in the presence of Cd2+ ions while sodium sulfide (Na2S) will remain soluble in the solution. Step 2: Filter out the cadmium sulfide (CdS) precipitate, which will separate the Cd2+ ions from the solution. The remaining solution will contain Na+ ions. Step 3: If necessary, evaporate the water from the remaining solution to obtain solid sodium ions in the form of a sodium salt, such as sodium chloride (NaCl).

To separate Cu2+ and Mg2+ ions from a solution mixture, we can make use of the difference in their complexation tendencies with ammonia (NH3). Step 1: Add excess ammonia solution to the solution containing Cu2+ and Mg2+ ions. This will cause the Cu2+ ions to form a soluble complex ion, [Cu(NH3)4]2+, while Mg2+ ions will remain as free ions in the solution. Step 2: Separate the [Cu(NH3)4]2+ complex ion from the Mg2+ ions by filtration, if there is a change in color or a formation of a precipitate. In some cases, you may need to use an ion-exchange column to perform this separation step. Step 3: If necessary, recover the Cu2+ and Mg2+ ions from their respective complexes and salts by evaporating the water or by performing a chemical reaction that releases the free metal ions.

To separate Pb2+ and Al3+ ions from a solution mixture, we can use the difference in solubility of their hydroxide salts. Step 1: Add a concentrated solution of sodium hydroxide (NaOH) to the solution containing Pb2+ and Al3+ ions. This will cause the formation of insoluble lead hydroxide, Pb(OH)2, while aluminum hydroxide, Al(OH)3, will remain soluble in the solution. Step 2: Filter out the lead hydroxide (Pb(OH)2) precipitate, which will separate the Pb2+ ions from the solution. The remaining solution will contain Al3+ ions. Step 3: If necessary, recover the Al3+ and Pb2+ ions from their respective hydroxide salts by performing a chemical reaction that releases the free metal ions, such as by acidification with hydrochloric acid (HCl).

To separate Ag+ and Hg2+ ions from a solution mixture, we can utilize the difference in solubility of their chloride salts. Step 1: Add a solution of sodium chloride (NaCl) to the solution containing Ag+ and Hg2+ ions. This will cause the formation of insoluble silver chloride (AgCl) and insoluble mercury(I) chloride (Hg2Cl2). Step 2: Filter out the silver chloride (AgCl) and mercury(I) chloride (Hg2Cl2) precipitates, which will separate the Ag+ and Hg2+ ions from the solution. Step 3: To separate AgCl from Hg2Cl2, add aqueous ammonia solution to the mixture of the two precipitates. This will cause the AgCl to dissolve into [Ag(NH3)2]+ complex ion, while Hg2Cl2 will remain insoluble. Filter out the insoluble Hg2Cl2. Step 4: If necessary, recover the Ag+ and Hg2+ ions from their respective chloride salts by performing a chemical reaction that releases the free metal ions, such as by acidification with nitric acid (HNO3).