Question

You have a solution that contains \(\mathrm{AgNO}_{3}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) Devise a separation method that results in having \(\mathrm{Ag}^{+}\) in one test tube, \(\mathrm{Pb}^{2+}\) in another, and \(\mathrm{Cu}^{2+}\) in a third test tube. Use solubility guidelines and \(K_{\mathrm{sp}}\) and \(K_{\mathrm{f}}\) values.

Short answer

Add NaCl to precipitate AgCl, KI for PbI2, and Na2S for CuS. Collect the precipitates separately.

Step-by-step solution

Understand the Solubility Rules

The first step is to recall the solubility rules to determine which compounds can form insoluble precipitates. According to solubility rules, silver chloride (AgCl) is insoluble, lead(II) iodide (PbI2) is insoluble, and copper(II) sulfide (CuS) is insoluble. This information will help us in separating the ions.

Select Appropriate Reagents

Given the solubility rules, add an appropriate reagent to precipitate each ion separately. To separate Ag+, add chloride ions (Cl-) to form AgCl, which will precipitate. To separate Pb2+, add iodide ions (I-) to form PbI2, which will precipitate. For Cu2+, add sulfide ions (S2-) to form CuS, which will precipitate.

Add Reagent for Silver Ion Separation

First, add a dilute solution of NaCl to the original solution. This step will result in the formation of a white precipitate of AgCl due to the reaction: \[\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl(s)}\] Filter off the AgCl precipitate and wash it for further purification if needed.

Add Reagent for Lead Ion Separation

Take the remaining solution after removing Ag+. Add a solution of potassium iodide (KI) to it. Pb2+ ions will form a yellow precipitate of PbI2 according to the reaction: \[\text{Pb}^{2+} + 2\text{I}^- \rightarrow \text{PbI}_2(s)\] Filter out the PbI2 precipitate and collect it in another test tube.

Add Reagent for Copper Ion Separation

To the solution left after removing Pb2+, add sodium sulfide (Na2S). This will precipitate Cu2+ as black copper sulfide (CuS) through the reaction: \[\text{Cu}^{2+} + \text{S}^{2-} \rightarrow \text{CuS(s)}\] Filter the CuS and transfer it into a separate test tube for storage.

Key concepts

Solubility Rules

When it comes to separating ions from a solution, solubility rules are your best friend. These rules help us understand which compounds can dissolve in water and which will form a solid known as a precipitate.
Understanding these rules is essential because they allow you to predict whether a compound will remain in a solution or fall out as a solid.
For example, a general solubility rule states that nitrates are soluble, but there are notable exceptions for chlorides and iodides. Like silver chloride (AgCl), which is insoluble in water. Here’s a quick list to remember:

• Most chloride (\( ext{Cl}^- \)) salts are soluble, except those of silver, lead, and mercury.
• Most iodide (\( ext{I}^- \)) salts are similarly soluble, with the same exceptions.
• Sulfides (\( ext{S}^{2-} \)) generally form insoluble compounds, with a few exceptions.

By using these rules, you can decide which ions will form a solid when you add specific reagents, aiding in their separation from a mixture.

Precipitation Reactions

Precipitation reactions are fascinating chemical processes that help us isolate certain ions from a solution. They are based on the formation of an insoluble product, termed a precipitate, when two aqueous solutions are mixed.
These reactions are driven by the solubility rules and the outcomes can often be predicted using these principles.
For example, to separate silver ions (\( ext{Ag}^+ \)) from a solution, you can add chloride ions (\( ext{Cl}^- \)).
This results in the formation of a white solid of silver chloride (\( ext{AgCl(s)} \)), a classic example of a precipitate:

• \( ext{Ag}^+ + ext{Cl}^- ightarrow ext{AgCl(s)} \)
• Lead ions (\( ext{Pb}^{2+} \)) can be precipitated by adding iodide ions (\( ext{I}^- \)), forming a yellow lead(II) iodide solid:\( ext{Pb}^{2+} + 2 ext{I}^- ightarrow ext{PbI}_2(s) \)
• Copper ions (\( ext{Cu}^{2+} \)) can be separated out by adding sulfide ions:\( ext{Cu}^{2+} + ext{S}^{2-} ightarrow ext{CuS(s)} \)

These simple reactions showcase how we can employ precipitation to effectively isolate and analyze different ions within a solution.

Ion Selectivity

Ion selectivity plays a crucial role in chemical separation techniques, such as in the exercise of separating ions from a solution.
This concept revolves around the meticulous choice of an ion or a group of ions that will selectively react to form a specific product, without affecting other ions in the mixture.
To achieve ion selectivity, you may consciously choose reagents that only interact with a particular ion due to its chemical properties. Consider selecting chloride ions (\( ext{Cl}^- \)) to precipitate out silver ions (\( ext{Ag}^+ \)), while leaving lead and copper ions still in solution:

\( ext{Ag}^+ \) specifically reacts with \( ext{Cl}^- \) to form \( ext{AgCl} \), easily isolating it from the mixture.

This principle is twofold: you get to pull one ion out specifically, and ensure your reacting agents do not disturb the other ions.
Selective precipitation in separation operations thus often revolves around exploiting physical or chemical differences, such as solubility, in the presence of multiple ions.Understanding these strategic selections allows chemists to efficiently conduct separations in a complex chemical world.