Question

Describe a simple test that would allow you to distinguish between \(\mathrm{AgNO}_{3}(s)\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\)

Short answer

Use hydrochloric acid to form a white precipitate with silver nitrate, confirming its presence.

Step-by-step solution

Understand the Composition of the Compounds

Firstly, understand that \(\mathrm{AgNO}_{3}(s)\) is solid silver nitrate and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\) is solid copper(II) nitrate. Silver nitrate contains silver ions (\(\mathrm{Ag}^{+}\)), while copper(II) nitrate contains copper ions (\(\mathrm{Cu}^{2+}\)).

Choose a Suitable Reagent

A simple test that can be used to identify these compounds is the reaction with chloride ions. When \(\mathrm{AgNO}_{3}(s)\) is dissolved in water and a chloride source like \(\mathrm{HCl}\) is added, a white precipitate of silver chloride \(\mathrm{AgCl}(s)\) forms. For \(\mathrm{Cu(NO}_{3}\)\(_{2})(s)\), the same test yields no precipitate as \(\mathrm{CuCl}_{2}\) is soluble in water.

Perform the Reaction

To distinguish between \(\mathrm{AgNO}_{3}(s)\) and \(\mathrm{Cu(NO}_{3}\)\(_{2}(s)\), dissolve a small amount of each compound separately in water. Add a dilute solution of \(\mathrm{HCl}\) to each solution. Observe the results.

Analyze the Results

For the silver nitrate solution, the appearance of a white precipitate of \(\mathrm{AgCl}(s)\) confirms the presence of \(\mathrm{AgNO}_{3}(s)\). No precipitate forms in the copper(II) nitrate solution, indicating it is \(\mathrm{Cu(NO}_{3}\)\(_{2}(s)\).

Conclude the Test

Based on the observed reactions, you can conclude which sample is silver nitrate (precipitate forms) and which is copper(II) nitrate (no precipitate).

Key concepts

Silver Nitrate

Silver nitrate is a crucial compound in chemical analysis because it contains silver ions (\(\text{Ag}^+\)) that are highly reactive.
In solid form, silver nitrate appears colorless or white.
When dissolved in water, the compound dissociates completely to release silver ions along with nitrate ions.
These ions are important in distinguishing silver nitrate from other chemical compounds.
In laboratory settings, silver nitrate is often used as a reagent in precipitation reactions, particularly due to its interaction with chloride ions.
Its key role in analytical chemistry lies in its ability to form a white, insoluble precipitate when it comes into contact with chloride ions.

• Commonly used in tests for chloride ions.
• Forms a dense white precipitate, \(\text{AgCl(s)}\).
• Helps identify and distinguish compounds through chemical reactions.

Copper(II) Nitrate

Copper(II) nitrate, symbolized as \(\text{Cu(NO}_3)_2\), is another important compound typically appearing as a blue solid due to its copper content.
It dissolves easily in water, and when in solution, it releases copper ions (\(\text{Cu}^{2+}\)) along with nitrate ions.
In contrast to silver nitrate, copper(II) nitrate does not react with chloride ions to form a precipitate.
This property is advantageous in testing and confirming the difference between it and silver nitrate, as reactions with chloride ions result in no visible precipitate.
The copper ions remain in solution, maintaining its blue hue.

• Releases \(\text{Cu}^{2+}\) ions when dissolved.
• Does not form a precipitate with chloride ions.
• Useful to compare against silver nitrate in precipitation tests.

Precipitation Reaction

Precipitation reactions are a type of chemical reaction where two solutions react, and an insoluble solid called a precipitate forms.
This is a crucial concept in identifying compounds through chemical analysis.
In our context, the reaction of silver nitrate with chloride ions serves as a textbook example of a precipitation reaction.
When silver nitrate is combined with a chloride source, such as hydrochloric acid (\(\text{HCl}\)), silver chloride, a white precipitate, forms.
Observing the formation of this precipitate is critical for determining the presence of silver ions.

• Involves insoluble solids appearing in chemical solutions.
• Helps differentiate compounds through visual confirmation.
• Common in analytical techniques to identify specific ions.

Chloride Ions

Chloride ions are important in the context of chemical tests and analyses.
Represented as \(\text{Cl}^-\), these ions are capable of reacting with various metal ions to form precipitates.
They are especially significant in distinguishing silver nitrate from other compounds like copper(II) nitrate through selective precipitation.
When chloride ions are introduced to a solution of silver nitrate, a visible white precipitate forms, indicating the formation of silver chloride.
This reaction does not occur with copper(II) nitrate, as chloride ions remain dissolved alongside the copper ions.

• Form white precipitates with silver ions.
• Key in conducting precipitation tests to identify certain ions.
• Does not precipitate with copper ions, remaining in solution.

Solubility

Solubility is a fundamental property that dictates how compounds interact in a liquid medium.
It refers to the maximum amount of a substance that can dissolve in a solvent at a given temperature.
In chemical analysis, understanding solubility helps predict and confirm outcomes of reactions, such as the precipitation of solids.
For example, silver chloride is not soluble in water, a property which is crucial in identification processes because it forms a precipitate in the presence of chloride ions.
Conversely, copper chloride forms a soluble compound, which means no precipitate is formed if copper(II) nitrate is tested using a source of chloride ions.

• Determines the formation of precipitates in reactions.
• Key to distinguishing between soluble and insoluble compounds.
• Guides predictability in chemical analysis outcomes.