Problem 80 - Chemistry The Central Science

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

Step-by-Step Solution

Answer

(a) Add \(\mathrm{H}_2\mathrm{S}\) to precipitate \(\mathrm{CdS}\), filter to separate \(\mathrm{Na}^+\) ions. (b) Adjust pH to 10 to precipitate \(\mathrm{Cu(OH)}_2\), filter to separate \(\mathrm{Mg}^{2+}\) ions. (c) Add \(\mathrm{HCl}\) to form \(\mathrm{PbCl}_2\) precipitate, filter to separate \(\mathrm{Al}^{3+}\) ions. (d) Add \(\mathrm{HCl}\), filter, rinse, add hot \(\mathrm{NH}_3\) to dissolve \(\mathrm{AgCl}\), filter to separate \(\mathrm{Ag}+\) ions and \(\mathrm{Hg}_2\mathrm{Cl}_2\).

Step 1: (a) Separating Na+ and Cd2+

To separate Na+ and Cd2+ ions from the mixture, we can use the following steps: 1. Add an excess amount of \(\mathrm{H}_2\mathrm{S}\) gas into the solution. This will precipitate \(\mathrm{CdS}\) but leave \(\mathrm{Na}^+\) ions in the solution since \(\mathrm{Na}^+\) is not a precipitate with \(\mathrm{H}_2\mathrm{S}\). 2. Filter the solution to remove the \(\mathrm{CdS}\) precipitate and collect the filtrate containing \(\mathrm{Na}^+\) ions.

Step 2: (b) Separating Cu2+ and Mg2+

To separate Cu2+ and Mg2+ ions from the mixture, we can follow these steps: 1. Adjust the pH of the solution to around 10 by adding \(\mathrm{NH}_3\) or \(\mathrm{NaOH}\). This will precipitate \(\mathrm{Cu(OH)}_2\) while the \(\mathrm{Mg}^{2+}\) ions will remain in the solution. 2. Filter the solution to separate the \(\mathrm{Cu(OH)}_2\) precipitate from the solution containing \(\mathrm{Mg}^{2+}\) ions.

Step 3: (c) Separating Pb2+ and Al3+

To separate Pb2+ and Al3+ ions from the mixture, follow these steps: 1. Add \(\mathrm{HCl}\) or \(\mathrm{NaCl}\) to the solution to form a precipitate of \(\mathrm{PbCl}_2\). The \(\mathrm{Al}^{3+}\) ions will not react, and hence will remain in the solution. 2. Filter the solution to separate the \(\mathrm{PbCl}_2\) precipitate and collect the supernatant containing \(\mathrm{Al}^{3+}\) ions.

Step 4: (d) Separating Ag+ and Hg2+

To separate Ag+ and Hg2+ ions from the mixture, follow this procedure: 1. Add \(\mathrm{HCl}\) or \(\mathrm{NaCl}\) to the solution. This will form a precipitate of \(\mathrm{AgCl}\) and \(\mathrm{Hg}_2\mathrm{Cl}_2\). 2. Filter the solution to separate the mixture of \(\mathrm{AgCl}\) and \(\mathrm{Hg}_2\mathrm{Cl}_2\) precipitates from the solution. 3. Rinse the mixture of precipitates with cold water to wash off any excess \(\mathrm{HCl}\) or \(\mathrm{NaCl}\). 4. Add a hot, concentrated solution of \(\mathrm{NH}_3\) to the mixture of precipitates. \(\mathrm{AgCl}\) will dissolve in \(\mathrm{NH}_3\) to form \(\mathrm{[Ag(NH}_3)_2]\mathrm{Cl}\) complex while \(\mathrm{Hg}_2\mathrm{Cl}_2\) will not dissolve since it is insoluble in \(\mathrm{NH}_3\). The resulting solution contains the \(\mathrm{Ag}+\) ions. 5. Filter the solution again to separate the supernatant containing the \(\mathrm{Ag}+\) ions from the undissolved \(\mathrm{Hg}_2\mathrm{Cl}_2\).

Key Concepts

Precipitation Reactions

Precipitation reactions are a fundamental chemical process used to separate ions in a solution by transforming them into an insoluble solid. This process involves mixing solutions of two soluble compounds, leading to the formation of an insoluble compound that precipitates out of the solution. In our guide to separating cations, the precipitation reactions play a crucial role. For instance, when separating

• Na⁺ and Cd²⁺,
• Cu²⁺ and Mg²⁺,
• Pb²⁺ and Al³⁺,
• Ag⁺ and Hg²⁺

from their mixtures, specific reagents are added to induce precipitation of one ion over the other. The insolubility of the resulting compound is key to this separation process.

For instance, adding H₂S gas to a solution containing Na⁺ and Cd²⁺ causes CdS to precipitate, thereby isolating Cd²⁺ from Na⁺. Precipitation reactions like these are guided by solubility rules, which predict whether a precipitate will form when two solutions are mixed.

Filtering Techniques

Once a precipitation reaction has taken place, separating the newly formed solid, or precipitate, from the liquid needs effective filtering techniques. Filtration allows the clear separation of the solid precipitate from the remaining liquid solution.

In the provided examples, filtration follows each precipitation step. To separate the solid, a filter paper is placed in a funnel, and the liquid is poured through it. The precipitate is left on the paper, while the clear filtrate passes through. This allows you to collect

• Cd²⁺ as CdS after separating it from Na⁺,
• Cu(OH)₂ from a solution after adjusting the pH, isolating it from Mg²⁺,
• PbCl₂ from Al³⁺ ions,
• AgCl from Hg₂Cl₂ when separating Ag⁺ and Hg²⁺.

This simple yet effective method ensures the precise collection of the desired cations or ions, making it a staple in chemistry labs for such separation tasks.

Solution pH Adjustment

Adjusting the pH of a solution is a critical step in altering the solubility of different ions. Many metal hydroxides are amphoteric, meaning that their solubility changes with pH, which one can exploit to selectively precipitate certain ions.

In the scenario of separating Cu²⁺ and Mg²⁺, altering the pH to around 10 induces the precipitation of Cu(OH)₂. This is because copper(II) hydroxide becomes insoluble as the pH is raised, while magnesium ions remain soluble at this pH. This method shows how adjusting pH can aid in targeted separation of cations

By understanding the hydroxides' solubility at different pH levels, chemists can efficiently cause selective ions to precipitate, allowing for their isolation through subsequent filtering.

Selective Precipitation

Selective precipitation is a powerful technique used to isolate specific ions from a mixture. By leveraging differences in solubility of compounds, one ion can be targeted for removal without disturbing others.

In the process of separating Ag⁺ and Hg²⁺, selective precipitation is employed by adding HCl. Both Ag⁺ and Hg₂²⁺ ions form chlorides, AgCl and Hg₂Cl₂, which are both initially precipitated out, requiring further distinction.

To specifically resolve this, silver chloride's distinct property of forming a soluble complex with ammonia is utilized. Adding concentrated NH₃ dissolves AgCl but leaves Hg₂Cl₂ intact. This exploitation of selective solubility differences helps in isolating one cation over another, a critical concept in practical ion separation tasks.