Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

Mixture \(\mathrm{X}=0.02 \mathrm{~mol}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution. 1 litre of mixture \(\mathrm{X}+\mathrm{excess} \mathrm{AgNO}_{3} \longrightarrow \mathrm{Y}\) 1 litre of mixture \(\mathrm{X}+\) excess \(\mathrm{BaCl}_{2} \longrightarrow \mathrm{Z}\) Number of moles of \(\mathrm{Y}\) and \(\mathrm{Z}\) are (a) \(0.02,0.01\) (b) \(0.01,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

Step-by-Step Solution

Answer

The number of moles of Y and Z are 0.01, 0.01.

Step 1: Understanding the Components of the Mixture

The mixture contains two compounds: \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}] \mathrm{Br}\) and \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}] \mathrm{SO}_{4}\). Each is present in the solution such that their total amounts to 0.02 mol, suggesting they are present in equal amounts, 0.01 mol each, since their sum is 0.02 mol.

Step 2: Reacting with Excess AgNO3

We react 1 litre of the mixture with excess AgNO3. The compound \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}] \mathrm{Br}\) will react with AgNO3 to form AgBr as \([\mathrm{Ag}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{AgBr}\downarrow]\). Since there are 0.01 mol of this compound, we get 0.01 mol of AgBr (compound Y). The other compound does not contribute to Y production in this reaction.

Step 3: Reacting with Excess BaCl2

We react 1 litre of the mixture with excess BaCl2. The compound \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}] \mathrm{SO}_{4}\) will react with BaCl2 to form BaSO4 as \([\mathrm{Ba}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{BaSO}_{4}\downarrow]\). Since there are 0.01 mol of this compound, we get 0.01 mol of BaSO4 (compound Z). The other compound does not contribute to Z production in this reaction.

Step 4: Analyzing Both Reactions

Both reactions are based on the principle that a precipitate forms only when the corresponding ion from the compound in the mixture reacts with the added reactant (AgNO3 or BaCl2). Therefore, the moles of the precipitates (Y and Z) both amount to 0.01 mol each, based on the amount of each complex compound that provided the respective ions: Br⁻ for Y and SO₄²⁻ for Z.

Key Concepts

Stoichiometry

Stoichiometry is the branch of chemistry that examines the quantitative relationships between reactants and products in chemical reactions. It allows us to predict how much of a particular reactant is needed or how much product will be formed in a reaction.
In the given exercise, stoichiometry helps us understand the amounts of compounds present in the mixture and the products formed. We know that the total amount of the two coordination compounds is 0.02 mol. Each compound contributes equally to this total, suggesting 0.01 mol each.

• Reacting \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) with AgNO\(_3\) yields AgBr. The stoichiometric ratio tells us that each mole of \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) releases one mole of Br⁻ ions to create an equal amount of AgBr precipitate.
• Similarly, \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) reacts with BaCl\(_2\) to form BaSO\(_4\), following a 1:1 stoichiometric ratio between SO₄²⁻ ions and BaCl\(_2\) to produce BaSO\(_4\).

The stoichiometric principles guide us to determine that 0.01 mol of each precipitate, Y (AgBr) and Z (BaSO\(_4\)), is formed.

Precipitation Reactions

Precipitation reactions occur when two soluble substances in solution react to form an insoluble product called a precipitate.
In this scenario, the precipitation reactions are specifically focused on how certain ions from the coordination compounds interact with the added reagents. When \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\) is mixed with AgNO\(_3\), Br⁻ ions combine with Ag⁺ ions to form AgBr, a classic example of a precipitation reaction.

• The formation of AgBr is indicated by the appearance of a solid in the otherwise clear solution, which settles out from the solution.
• Similarly, when \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) interacts with BaCl\(_2\), SO₄²⁻ ions pair with Ba²⁺ ions to produce BaSO\(_4\) as a solid precipitate.

These reactions highlight the selective precipitation ability of various ions in solution, providing a visual confirmation of the reactions occurring.

Complex Ion Equilibria

Complex ion equilibria describe the balance between different forms of metal complexes in solution. Coordination compounds like those in the exercise consist of a central metal ion surrounded by ligands.
Each complex can dissociate in solution to some degree, releasing its constituent ions. For \([\text{Co}(\text{NH}_3)_5 \text{SO}_4] \text{Br}\), when the solution is mixed with AgNO\(_3\), the Br⁻ ions become available to react with Ag⁺ ions.
Meanwhile, \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) can release SO₄²⁻ ions when mixed with BaCl\(_2\), allowing them to interact with Ba²⁺ ions.

• The equilibrium state in solutions containing these coordination compounds dictates the availability of free ions for reaction.
• When a precipitation reaction occurs, the removal of ions from the solution shifts the equilibrium, often allowing more ions to dissociate to maintain the equilibrium.

Understanding complex ion equilibria enables chemists to predict the concentrations of ions at equilibrium and their subsequent reactions in solution.