Problem 95 - A Complete Resource Book in Chemistry for JEE Main

Question

The only cations present in a slightly acidic solution are \(\mathrm{Fe}^{3+}, \mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\). The reagent that when added in excess of this solution would identify and separate \(\mathrm{Fe}^{3+}\) in one step is (a) \(2 \mathrm{M} \mathrm{HCl}\) (b) \(6 \mathrm{M} \mathrm{NaOH}\) (c) \(6 \mathrm{M} \mathrm{NH}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{~S}\) gas

Step-by-Step Solution

Answer

The answer is (b) \(6 \mathrm{M} \mathrm{NaOH}\).

Step 1: Analyze the Reagents

We need to determine which reagent selectively reacts with \(\mathrm{Fe}^{3+}\) and forms a precipitate. \(\mathrm{Fe}^{3+}\) can form distinct complexes or precipitates under different conditions. Let's analyze each reagent one by one.

Step 2: Test \(2 \mathrm{M} \mathrm{HCl}\)

Adding \(2 \mathrm{M} \mathrm{HCl}\) will increase the acidity of the solution. \(\mathrm{Fe}^{3+}\) could form \(\mathrm{FeCl}_4^-\) in a highly acidic solution, but increasing the acidity does not selectively precipitate \(\mathrm{Fe}^{3+}\). It does not separate \(\mathrm{Fe}^{3+}\) in one step.

Step 3: Test \(6 \mathrm{M} \mathrm{NaOH}\)

Adding \(6 \mathrm{M} \mathrm{NaOH}\) would create a basic environment, converting \(\mathrm{Fe}^{3+}\) into \(\mathrm{Fe(OH)}_3\), which is insoluble and forms a precipitate. This reaction specifically precipitates \(\mathrm{Fe}^{3+}\) and distinguishes it from \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\), as their hydroxides are soluble in excess hydroxide.

Step 4: Test \(6 \mathrm{M} \mathrm{NH}_3\)

Adding \(6 \mathrm{M} \mathrm{NH}_3\) forms \(\mathrm{Fe(OH)}_3\) precipitate in excess ammonia, but \(\mathrm{Cu}^{2+}\) can also form a complex \(\mathrm{Cu(NH}_3)_4^{2+}\), complicating the separation. \(\mathrm{NH}_3\) is less effective in separating \(\mathrm{Fe}^{3+}\) in one step.

Step 5: Test \(\mathrm{H}_{2} \mathrm{~S}\) gas

Passing \(\mathrm{H}_{2} \mathrm{~S}\) through the solution could form sulfide precipitates, but \(\mathrm{Fe}^{3+}\), \(\mathrm{Zn}^{2+}\), and \(\mathrm{Cu}^{2+}\) can all form sulfide precipitates under different conditions. This does not effectively separate \(\mathrm{Fe}^{3+}\).

Step 6: Conclusion: Choose the Correct Reagent

The best reagent that can be added to separate \(\mathrm{Fe}^{3+}\) in one step is \(6 \mathrm{M} \mathrm{NaOH}\). It precipitates \(\mathrm{Fe(OH)}_3\) while leaving \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\) in solution.

Key Concepts

Cations in Acidic Solution

When you have cations like

• \( \mathrm{Fe}^{3+} \)
• \( \mathrm{Zn}^{2+} \)
• \( \mathrm{Cu}^{2+} \)

in a slightly acidic solution, you're dealing with positively charged ions in an environment where there is more concentration of hydrogen ions (H\(^+\)).

In such conditions, these cations are less likely to react because the acidity stabilizes them. This environment is important because it influences how these ions will interact with different reagents.

For instance, excess hydrogen ions from the acidity might prevent certain reactions from occurring, as they can compete with the cations for the reagents. This makes it crucial to select a reagent that can provoke a selective reaction with only one of the cations.

Precipitation Reactions

A precipitation reaction occurs when two soluble solutions are mixed and an insoluble solid forms. This solid is called a precipitate.

In our exercise, we need a reagent that can react with

• \( \mathrm{Fe}^{3+} \)

to form a precipitate, effectively separating it from

• \( \mathrm{Zn}^{2+} \)
• \( \mathrm{Cu}^{2+} \)

An example is the reaction of \( \mathrm{Fe}^{3+} \) with hydroxide ions

This leads to the formation of iron(III) hydroxide, \( \mathrm{Fe(OH)}_3 \), a precipitate that is insoluble in water. This type of reaction is useful in analytical chemistry for the determination and separation of ions.

Metal Hydroxides

Metal hydroxides are compounds containing metal ions and hydroxide ions (OH\(^-\)).

In a basic solution, many metal ions can react with excess hydroxide ions to form metal hydroxides.

For example, when you add \( 6\, \mathrm{M} \; \mathrm{NaOH} \) in the solution containing

• \( \mathrm{Fe}^{3+} \)

it reacts to form \( \mathrm{Fe(OH)}_3 \). This causes

• \( \mathrm{Fe}^{3+} \)

to precipitate, making it easy to separate from the solution.

This is especially helpful since other ions like

• \( \mathrm{Zn}^{2+} \)
• \( \mathrm{Cu}^{2+} \)

remain soluble, allowing a simple separation by filtration.

Selective Precipitation

Selective precipitation is a technique used to separate specific ions from a solution.

This is achieved by adding a reagent that will form a precipitate with one specific ion, leaving others in solution. This concept is key when multiple cations are present.One successful example from our exercise was using \( 6\, \mathrm{M} \; \mathrm{NaOH} \) and its ability to selectively precipitate

• \( \mathrm{Fe}^{3+} \)

ions as \( \mathrm{Fe(OH)}_3 \). Since \( \mathrm{Zn}^{2+} \) and \( \mathrm{Cu}^{2+} \) hydroxides remain dissolved in the excess of hydroxide ions, a simple filtration can then remove the iron precipitate.

This method is advantageous as it allows for the targeted separation of specific ions in a multicomponent solution.