Question

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 \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+},(\mathbf{d}) \mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+}\).

Short answer

To separate the given cation mixtures: a) Add sodium sulfide to the Na⁺ and Cd²⁺ mixture, form a precipitate of CdS, and then filter the mixture to obtain Na²S and CdS. b) Add excess ammonia to the Cu²⁺ and Mg²⁺ mixture, form a copper-ammonia complex, recover Mg²⁺ ions using acid, and then add excess acid to recover the copper ions. c) Adjust pH to 10 for the Pb²⁺ and Al³⁺ mixture using NaOH, form a precipitate of Pb(OH)₂, filter the solution, dissolve the precipitate in acid, and then recover aluminum ions using acid. d) Add sodium chloride to the Ag⁺ and Hg²⁺ mixture, form a precipitate of AgCl, filter the mixture to obtain AgCl and Hg²⁺ ions, recover silver ions by dissolving AgCl in ammonia, and evaporate the filtrate to obtain solid mercuric chloride.

Step-by-step solution

a) Separation of \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+}\) ions

To separate the sodium and cadmium ions, we can take advantage of the fact that cadmium ions can form insoluble salts with some anions, while sodium salts are generally soluble. 1. Add an excess of sodium sulfide (\(\mathrm{Na}_2\mathrm{S}\)) solution to the mixture. This will form a precipitate of cadmium sulfide \(\mathrm{(CdS)}\) as it is insoluble in water: \[ \mathrm{Cd}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{CdS} \] 2. Filter the resulting mixture. The solid precipitate containing \(\mathrm{CdS}\) will be collected on the filter paper, and the filtrate will contain the \(\mathrm{Na}^+\) ions in the form of sodium sulfide. 3. Wash and dry the precipitate to obtain pure cadmium sulfide. Evaporate the filtrate to obtain solid sodium sulfide.

b) Separation of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions

To separate copper and magnesium ions, we can use ammonia solution as a complexing agent. 1. Add an excess of ammonia solution to the mixture. Copper ions will form a deep blue complex with ammonia, while magnesium ions do not form complexes and remain as \(\mathrm{Mg}^{2+}\) in the solution. \[ \mathrm{Cu}^{2+} + 4\mathrm{NH}_3\rightarrow [\mathrm{Cu(\mathrm{NH}_3)_4}]^{2+} \] 2. Filter the resulting mixture. The deep blue solution containing the copper-ammonia complex will pass through the filter, and the \(\mathrm{Mg}^{2+}\) ions will stay in the residue. 3. The residue can be treated with an acid (e.g. hydrochloric acid) to release the magnesium ions back into the solution. Magnesium salt can be recovered by evaporating the solvent. 4. To recover copper ions, add excess acid (e.g. sulfuric acid) to the blue filtrate, which will displace the ammonia and release the copper ions. Copper salt can be recovered by evaporating the solvent.

c) Separation of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\) ions

To separate lead and aluminum ions, we can take advantage of their solubility differences at different pH levels. 1. Add a small amount of sodium hydroxide (\(\mathrm{NaOH}\)) solution dropwise to the mixture until the pH is around 10. At this pH, lead ions will form a precipitate of lead hydroxide, while aluminum ions remain in solution as soluble complexes with hydroxide ions. \[ \mathrm{Pb}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Pb(OH)_2} \] 2. Filter the solution. The precipitate containing \(\mathrm{Pb(OH)_2}\) will be collected on the filter paper, and the filtrate will contain the aluminum ions as soluble complexes. 3. To obtain the lead salt, dissolve the lead hydroxide precipitate in an acid (e.g. nitric acid) and evaporate the solvent. 4. To recover aluminum ions, add acid (e.g. hydrochloric acid) to the filtrate to displace the hydroxide ions and convert aluminum back to its ionic form. Aluminum salt can be recovered by evaporating the solvent.

d) Separation of \(\mathrm{Ag}^+\) and \(\mathrm{Hg}^{2+}\) ions

To separate silver and mercury ions, we can take advantage of the fact that silver forms an insoluble chloride while mercuric chloride is soluble. 1. Add a solution of sodium chloride (\(\mathrm{NaCl}\)) to the mixture. Silver ions will form a precipitate of silver chloride, while mercury ions will remain in solution as soluble mercuric chloride. \[ \mathrm{Ag}^+ + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl} \] 2. Filter the resulting mixture. The precipitate containing \(\mathrm{AgCl}\) will be collected on the filter paper, and the filtrate will contain the \(\mathrm{Hg}^{2+}\) ions. 3. Wash and dry the precipitate to obtain pure silver chloride. To recover silver ions, dissolve the silver chloride in an appropriate solution (e.g. ammonia) and evaporate the solvent. 4. To recover mercury ions, evaporate the filtrate to obtain solid mercuric chloride (be cautious during this step as mercury compounds are toxic).

Key concepts

Chemical Precipitation

Chemical precipitation is a fundamental technique in analytical chemistry used for separating cations in a mixture. By adding a reagent that forms an insoluble compound with a specific ion, a precipitate is created. For instance, in the separation of \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+}\), sodium sulfide is added to form cadmium sulfide \(\mathrm{CdS}\), which is insoluble in water. This process hinges on the principle that different ions have different solubilities with specific anions.

To optimize the success of precipitation, it is crucial to use an excess of the precipitating agent to ensure the complete reaction of the targeted cation. After precipitation, the mixture can be filtered to separate the solid from the liquid phase, isolating the desired compound. Remember, during the drying process, care must be taken to avoid contamination or loss of the product.

Key Points to Remember in Chemical Precipitation:

• Identify the specific ion to be precipitated.
• Select the appropriate reagent that will form an insoluble compound with that ion.
• Ensure an excess of the reagent to drive the precipitation to completion.
• Meticulously carry out the separation to avoid losing material.

Complexation Reactions

Complexation reactions involve the formation of a compound, referred to as a complex, in which a central metal atom or ion is bonded to one or more molecules or ions. During the separation of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions, ammonia acts as a complexing agent. Copper ions react with the ammonia to form a deep blue complex, whereas magnesium ions do not form such a complex and remain in solution. This difference allows for an easy separation through a process called selective complexation.

Complexation is particularly useful because it can often be reversed. By changing the conditions of the solution, such as its pH or by adding another reagent, the complex can break apart, releasing the metal ion. This reversibility aids in the recovery of the original cations from their complexes for further analysis or use.

Essentials of Complexation:

• Complex formation is highly specific to certain metal ions.
• Ammonia is a common complexing agent for copper ions.
• The ability to reverse the complexation offers an avenue for the recovery of the metal ion.

Solubility Differences

Solubility differences allow for the separation of ions based on the varying solubilities of their compounds. Not all ionic compounds dissolve equally in water, and by exploiting this fact, specific ions can be selectively precipitated out from a solution containing a mixture of ions. For example, in the separation of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\), the different solubilities of their hydroxide compounds in water are used to isolate them.

When \(\mathrm{NaOH}\) is added, lead precipitates as \(\mathrm{Pb(OH)_2}\), while aluminum forms soluble complexes. The careful selection of the precipitating agent and control of the reaction conditions, such as concentration and temperature, are essential for successful separation.

Crucial Points in Solubility-Based Separation:

• The solubility of ionic compounds varies with temperature and the nature of the solvent.
• Precise control of reaction conditions can lead to successful targeted precipitation.

pH Dependent Solubility

The solubility of many compounds, especially those of metal hydroxides, is pH dependent. In analytical chemistry, adjusting the pH of the solution can lead to the precipitation of specific ions while others remain in solution. This is particularly illustrated in the step where \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\) ions are separated. By adjusting the pH to around 10 using \(\mathrm{NaOH}\), lead ions are made to precipitate as lead hydroxide. In contrast, aluminum ions remain soluble at this pH due to the formation of soluble hydroxide complexes.

In practice, the process usually involves slow and controlled addition of an acid or a base to reach the desired pH. This gradual change aids in the formation of a purer precipitate and prevents overshooting the pH, which can affect the accuracy of the separation.

Considerations for pH Dependent Solubility:

• Understanding the pH at which a particular ion precipitates is essential.
• Careful, controlled addition of acid or base is necessary to achieve the desired pH.
• pH meters, indicators, or pH paper can be used for precise pH monitoring during the process.