Problem 126 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The answer is (a) 0.01, 0.01.

Step 1: Initial Information

We are given a mixture of two different compounds: \([\mathrm{Co}(\mathrm{NH}_{3})_{5} \mathrm{SO}_{4}] \mathrm{Br}\) and \([\mathrm{Co}(\mathrm{NH}_{3})_{5} \mathrm{Br}] \mathrm{SO}_{4}\), each with 0.02 moles. The mixture is diluted in a 2-liter solution.

Step 2: Concentration Calculation

For each compound, since we have 0.02 moles in a 2-liter solution, the concentration of each compound in the mixture is \( \frac{0.02}{2} = 0.01 \) moles per liter.

Step 3: Reaction with AgNO3

\([\mathrm{Co}(\mathrm{NH}_{3})_{5} \mathrm{SO}_{4}] \mathrm{Br}\) reacts with excess \(\mathrm{AgNO}_{3}\) to produce AgBr (a precipitate). With a concentration of 0.01 M and 1 liter used, 0.01 moles of AgBr are produced because 1 mole of the compound produces 1 mole of AgBr.

Step 4: Reaction with BaCl2

\([\mathrm{Co}(\mathrm{NH}_{3})_{5} \mathrm{Br}] \mathrm{SO}_{4}\) reacts with excess \(\mathrm{BaCl}_{2}\) to produce \(\mathrm{BaSO}_{4}\) (a precipitate). Similarly, using 1 liter with a concentration of 0.01 M, this yields 0.01 moles of \(\mathrm{BaSO}_{4}\) since 1 mole of the compound produces 1 mole of \(\mathrm{BaSO}_{4}\).

Step 5: Final Result

Combining calculations, both \(0.01\) mole of \(\mathrm{AgBr}\) and \(0.01\) mole of \(\mathrm{BaSO}_{4}\) are produced. Thus, the correct answer choice is option (a).

Key Concepts

Mole Concept

The Mole Concept plays a fundamental role in chemistry, serving as a bridge between the atomic world and tangible, measurable quantities. One mole is defined as exactly 6.022 x 10\^23 entities, whether they be atoms, molecules, or ions. This constant, known as Avogadro's number, allows chemists to count particles by measuring mass. In the original exercise, the mole concept is essential for determining the amount of reactant used in each reaction.
To calculate the concentration of each compound in solutions, you need to understand that a solution's molarity (M) is the number of moles of solute per liter of solution. With 0.02 moles of each compound diluted in a 2-liter solution, the concentration of each becomes:

• Concentration = \( \frac{0.02 \, \text{moles}}{2 \, \text{liters}} = 0.01 \, \text{M} \)

This calculation is critical for then predicting the outcomes in the subsequent precipitation reactions.

Precipitation Reaction

Precipitation reactions are chemical reactions where two solutions are mixed, resulting in the formation of an insoluble solid, called a precipitate. This occurs when the product of the reaction has very low solubility in water. In the context of the provided exercise, the reactions involve the formation of AgBr and BaSO\(_4\) as precipitates.
When

• \( [\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4] \mathrm{Br} \)

reacts with excess AgNO\(_3\), silver bromide (AgBr) precipitates. Each mole of this compound reacts stoichiometrically with one mole of AgNO\(_3\) to produce one mole of AgBr. Similarly, when

• \( [\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}] \mathrm{SO}_4 \)

reacts with excess BaCl\(_2\), barium sulfate (BaSO\(_4\)) is formed. Understanding the stoichiometry of these reactions is crucial. Since both compounds in the solution have a concentration of 0.01 M, reactions

• produce 0.01 moles of AgBr
• and 0.01 moles of BaSO\(_4\)

by engaging the _entire_ solution.

Chemical Equilibrium

Chemical equilibrium describes a state where the concentrations of reactants and products in a chemical reaction remain unchanged over time, indicating a balance in the rates of the forward and reverse reactions. However, it is important to note that equilibrium is dynamic - reactions are ongoing and simultaneous, but their effects cancel each other out.
In the case of precipitation reactions, equilibrium can influence solubility. However, in the context of the exercise, the notion of equilibrium isn't reached because the quantities of the reagents and conditions (like excess of reactants) favor complete precipitation. In other words, since one reactant is in excess and the other is completely consumed, the reaction proceeds to completion rather than equilibrium.
The only consideration of equilibrium in this context might arise if we were dealing with partially soluble compounds or if the experiment was designed to investigate the solubility product or K\(_{sp}\). Nonetheless, understanding equilibrium helps in determining when a reaction will stop progressing and can also aid in calculating yields and determining whether more product can be formed under different conditions.