Problem 39 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The volume (in \(\mathrm{mL}\) ) of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) required for complete precipitation of chloride ions present in \(30 \mathrm{~mL}\) of \(0.01 \mathrm{M}\) solution of \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\), as silver chloride is close to

Step-by-Step Solution

Answer

The required volume of AgNO3 is 6 mL.

Step 1: Determine the Moles of Chloride Ions

The compound \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}] \text{Cl}_2\) consists of 2 moles of Cl\(^{-}\) ions per mole of compound. We start by finding the moles of the compound: \(0.01 \, \text{M} \times 30 \, \text{mL} = 0.3 \, \text{mmol}\). Since each mole yields 2 moles of chloride ions, the moles of chloride ions are \(0.3 \, \text{mmol} \times 2 = 0.6 \, \text{mmol}\).

Step 2: Use Stoichiometry to Determine Required AgNO3

AgNO\(_3\) reacts with Cl\(^{-}\) ions to form AgCl according to the reaction: \(\text{AgNO}_3 + \text{Cl}^- \rightarrow \text{AgCl} + \text{NO}_3^-\). Thus, 1 mole of Cl\(^{-}\) reacts with 1 mole of AgNO\(_3\). From Step 1, we have \(0.6 \, \text{mmol}\) of Cl\(^{-}\) ions, so we need \(0.6 \, \text{mmol}\) of AgNO\(_3\).

Step 3: Calculate the Volume of AgNO3 Solution

Since the concentration of AgNO\(_3\) is \(0.1 \, \text{M}\), this means \(0.1 \, \text{mol/L} = 0.1 \, \text{mmol/mL}\). To find the required volume in \(\text{mL}\), use the formula: \[ V(\text{mL}) \times 0.1 \, \text{mmol/mL} = 0.6 \, \text{mmol} \]. Solving for \(V\), we find \( V = \frac{0.6}{0.1} = 6 \, \text{mL}\).

Key Concepts

Stoichiometry

Stoichiometry is a crucial aspect of understanding chemical reactions. It involves calculating the exact amounts of reactants and products in a chemical equation based on their molar relationships. Understanding stoichiometry helps to predict how much of a substance is required or produced in a given reaction.In our exercise, we see stoichiometry at work with the precipitation reaction:

• The compound \([ ext{Cr}( ext{H}_2 ext{O})_5 ext{Cl}] ext{Cl}_2\) contains two chloride ions per molecule.
• This ionic compound dissociates in solution, yielding twice the number of chloride ions as the number of moles of the compound itself.

Knowing this, we can calculate the moles of chloride ions to better understand the stoichiometric relationships in the reaction.
In this context, stoichiometry aids in determining the necessary amount of \( ext{AgNO}_3 \), ensuring the complete precipitation of these chloride ions to form \( ext{AgCl} \). By applying this concept, we estimate the precise volume needed to achieve our reaction goals.

Molarity

Molarity is one of the most common methods of expressing concentration, especially in chemistry. Defined as the number of moles of solute per liter of solution, molarity provides a way to quantify the concentration of a solution.In our example, molarity helps specify how concentrated the \( ext{AgNO}_3 \) and chloride solutions are:

• The \( ext{AgNO}_3 \) solution is given as \(0.1 \, ext{M}\), which means there are 0.1 moles of \( ext{AgNO}_3 \) per liter.
• The solution \( ext{[ ext{Cr}( ext{H}_2 ext{O})_5 ext{Cl}] ext{Cl}_2} \) is \(0.01 \text{M}\), signifying 0.01 moles per liter.

By applying molarity concepts, we can accurately calculate the volume of one solution needed to react with a specified amount of another. For example, using the concentration of \( ext{AgNO}_3 \), we can find out how many milliliters are necessary for complete precipitation, which is essential for practical laboratory applications and theoretical calculations.

Precipitation Reactions

Precipitation reactions are a type of reaction where two soluble salts in solution react to form an insoluble salt, called the precipitate. This process is significant in various fields, from analytical chemistry to environmental science, where identifying or removing ions is crucial.In the provided exercise:

• The \( ext{AgNO}_3 \) interaction with chloride ions forms silver chloride \( ext{AgCl} \), a classic example of a precipitation reaction.
• This solid precipitate indicates the presence of chloride ions in a solution, often used in titrations to determine their concentration.

The equation \( ext{AgNO}_3 + ext{Cl}^- ightarrow ext{AgCl} + ext{NO}_3^-\) demonstrates this clearly. Here, the insoluble \( ext{AgCl} \) forms, while \( ext{NO}_3^- \) remains in solution, often playing no direct role in the visible outcomes. Precipitation reactions like this one are vital for isolating substances or analyzing chemical composition, serving as an essential tool in both academic and practical chemistry settings.