Problem 53 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The volume (in \(\mathrm{mL}\) ) of \(0.125 \mathrm{M} \mathrm{AgNO}_{3}\) required to quantitatively precipitate chloride ions in \(0.3 \mathrm{~g}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is [Main Jan. \(\mathbf{0 8}, \mathbf{2 0 2 0}\) (I)] \(\mathrm{M}_{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{j} \mathrm{Cl}_{3}\right.}=267.46 \mathrm{~g} / \mathrm{mol}\) \(\mathrm{M}_{\mathrm{AgNO}_{3}}=169.87 \mathrm{~g} / \mathrm{mol}\)

Step-by-Step Solution

Answer

26.93 mL of 0.125 M AgNO3 is required.

Step 1: Understand the Problem

We need to find the volume (in \(\mathrm{mL}\)) of \(0.125\, \mathrm{M}\) \(\mathrm{AgNO}_{3}\) needed to completely precipitate the chloride ions in \(0.3 \, \mathrm{g}\) of the compound \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\). The molar mass of the compound is \(267.46 \, \mathrm{g/mol}\), and for \(\mathrm{AgNO}_{3}\), it is \(169.87 \, \mathrm{g/mol}\).

Step 2: Determine Moles of Compound

First, calculate the moles of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) using its mass and molar mass. \(\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.3 \, \mathrm{g}}{267.46 \, \mathrm{g/mol}} = 0.001122 \, \mathrm{mol}\).

Step 3: Determine Moles of Chloride Ions

Each molecule of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) contains 3 chloride ions. Thus, the moles of chloride ions are \(3 \times 0.001122 \, \mathrm{mol} = 0.003366 \, \mathrm{mol}\).

Step 4: Use Stoichiometry for Precipitation Reaction

In the reaction \(\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl} (s)\), each mole of chloride requires one mole of \(\mathrm{Ag}^{+}\) to form an insoluble precipitate of \(\mathrm{AgCl}\). Therefore, \(0.003366\) mol of \(\mathrm{Ag}^{+}\) is needed.

Step 5: Calculate Volume of \(\mathrm{AgNO}_{3}\) Solution

Using the molarity formula, \(\mathrm{M} = \frac{\mathrm{moles}}{\mathrm{volume}}\), rearrange to find the volume: \(\text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.003366 \, \mathrm{mol}}{0.125 \, \mathrm{mol/L}} = 0.026928 \, \mathrm{L}\). Converting to \(\mathrm{mL}\), we get \(0.026928 \, \mathrm{L} \times 1000 \approx 26.93 \, \mathrm{mL}\).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. In simple terms, it helps us understand how much of each substance is used or produced.
In the exercise we are looking at, stoichiometry is the backbone that allows us to determine how much silver nitrate (\(\mathrm{AgNO}_3\)) is needed to react completely with the chloride ions from the given compound. We start by calculating the moles of the compound, \([\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3\), using the equation:

• Moles of substance = \(\frac{\text{mass of substance}}{\text{molar mass of substance}}\)

This calculation gives us the moles of the entire compound. Next, we determine how these relate to the moles of chloride ions, as each unit of the compound contains three chloride ions. This step is crucial because stoichiometry maintains the balance between the number of moles of reactants and products as defined by the balanced chemical equation.
Finally, using stoichiometric principles, we equate the moles of \(\mathrm{AgNO}_3\) to the moles of chloride ions, since they react in a 1:1 ratio to form \(\mathrm{AgCl}\) precipitate.

Molarity

Molarity is a measurement of the concentration of a solution, expressed in terms of the amount of solute (in moles) per liter of solution. It's a key tool for chemists to prepare solutions with precise concentrations for reactions.
In the given problem, the molarity (\mathrm{M}) of \(\mathrm{AgNO}_3\) is provided as 0.125 \(\mathrm{mol/L}\). Knowing this allows us to calculate the volume of \(\mathrm{AgNO}_3\) solution needed for the reaction. The relationship between molarity, moles of solute, and volume of solution is given by the formula:

• \(\mathrm{M} = \frac{\mathrm{moles}}{\mathrm{volume \ in \ liters}}\)

With this equation, we can rearrange to solve for the volume when the number of moles is known:

• Volume \(\mathrm{(L)} = \frac{\mathrm{moles}}{\mathrm{M}}\)

This calculation tells us exactly how much \(\mathrm{AgNO}_3\) solution is needed in liters, which can be converted easily into milliliters for practical use in the lab. Understanding molarity ensures precise and accurate measurements in chemical reactions, essential for achieving the desired outcomes.

Precipitation Reactions

Precipitation reactions occur when two solutions react to form an insoluble solid, called a precipitate. This type of reaction is common in aqueous chemistry, where ions in solution combine to form a compound that does not dissolve in water. In the exercise scenario, the precipitation reaction is:

• \(\mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl}(s)\)

Here, silver ions (\mathrm{Ag}^{+}) from the silver nitrate solution react with chloride ions (\mathrm{Cl}^{-}) from the \([\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3\) compound to form solid silver chloride (\mathrm{AgCl}), a white precipitate.
The formation of a precipitate is a visual indication that a reaction has occurred. For chemists, these reactions are significant because they allow for the isolation and removal of specific ions from a solution. Understanding the process and calculations involved in precipitation reactions is crucial to predicting the products formed and their quantities.
Molecules in precipitation reactions rely on stoichiometric calculations to ensure all chloride ions are captured as \(\mathrm{AgCl}\), demonstrating the importance of precise measurements and calculations in chemical reactions.