Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(0.001\) molal solution of \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{4}\right]\) in water had a freezing point depression, \(0.0056^{\circ} \mathrm{C} . \mathrm{K}_{\mathrm{f}}\) for water is \(1.86^{\circ} \mathrm{cm}^{-1} .\) The number of moles of \(\mathrm{AgNO}_{3}\) required to react with one mole of the complex in aqueous solution is

Step-by-Step Solution

Answer

2 moles of \( \mathrm{AgNO}_{3} \) are required.

Step 1: Understanding Freezing Point Depression

Freezing point depression is a colligative property, which depends on the number of particles in solution. It is calculated using \[ \Delta T_f = i \times K_f \times m \]where \( \Delta T_f = 0.0056^{\circ} \mathrm{C} \), \( K_f = 1.86^{\circ} \mathrm{C \cdot m}^{-1} \), and \( m = 0.001 \). The van't Hoff factor \( i \) represents the number of particles the compound dissociates into in solution.

Step 2: Calculate the Van't Hoff Factor

Rearrange the freezing point depression formula to solve for \( i \): \[ i = \frac{\Delta T_f}{K_f \times m} = \frac{0.0056}{1.86 \times 0.001} = 3 \]. This indicates that the complex dissociates into 3 particles in solution.

Step 3: Determine the Dissociation

The complex \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}]\) dissociates in water. Based on \( i = 3 \), one possible dissociation pathway is:\[ [\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}] \rightarrow [\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}]^{2+} + 2\mathrm{Cl}^- \]The complexes and dissociation products align with the van't Hoff factor.

Step 4: Calculate AgNO3 Required

To react with one mole of the complex, 2 moles of \( \mathrm{AgNO}_{3} \) are needed to precipitate the 2 moles of chloride ions as \( \mathrm{AgCl} \) using the reaction:\[ 2 \mathrm{AgNO}_{3} + 2\mathrm{Cl}^- \rightarrow 2 \mathrm{AgCl} + 2 \mathrm{NO}_{3}^- \].Therefore, 2 moles of \( \mathrm{AgNO}_{3} \) are required to fully react with one mole of \( [\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}] \).

Key Concepts

Freezing Point Depression

Freezing point depression is an interesting phenomenon where the freezing point of a solvent is lowered when a solute is added. This is a colligative property, meaning it depends primarily on the number of dissolved particles, not their identity. When a non-volatile solute is added to a solvent like water, the liquid's freezing point drops.
In the given exercise, the freezing point depression formula, \(\Delta T_f = i \times K_f \times m\), helps determine how much the freezing point is lowered. Here, \(\Delta T_f\) is the depression in temperature, \(K_f\) is the cryoscopic constant of the solvent, and \(m\) is the molality of the solution. This principle allows us to explore how different compounds behave in solution by affecting the freezing point.

Van't Hoff Factor

The van't Hoff factor, denoted by \(i\), is crucial in understanding how solutes affect colligative properties. It signifies the number of particles into which a solute dissociates in solution. In simple terms, \(i\) tells us about the ionization or association happening when a compound dissolves.
In the instance of \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}]\), calculating \(i\) from the formula \(i = \frac{\Delta T_f}{K_f \times m}\) yielded a van't Hoff factor of 3. This means the compound dissociates into three particles, influencing how much the freezing point is depressed. It's important to note how this factor can vary depending on whether the solute is ionic or covalent.

Chemical Dissociation

Chemical dissociation is when compounds split into smaller entities like ions when dissolved. This is important as it directly impacts colligative properties by altering the number of solute particles in the solution.
For the compound \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}]\), it dissociates in water, forming \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}]^{2+}\) and two chloride ions (\(\mathrm{Cl}^-\)). This aligns with the van't Hoff factor of 3, confirming that the substance splits into three components: one positive ion and two negative ions. Understanding this division helps us comprehend the overall effect on solution properties.

Molality

Molality is a measure of the concentration of a solute in a solution, particularly used in calculating colligative properties. It is defined as the moles of solute per kilogram of solvent, denoted by \(m\). This unit remains unaffected by temperature changes, making it ideal for calculations involving freezing point depression and boiling point elevation.
In the exercise, a 0.001 molal solution was referenced, meaning there are 0.001 moles of \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}]\) per kilogram of water. The small concentration hints at only a slight change in the solvent's freezing point, pivotal in observing precise changes described by colligative properties.

Inorganic Chemistry

Inorganic chemistry focuses on compounds that are not primarily based on carbon-hydrogen bonds, comprising most of the periodic table elements. It encompasses a wide range of substances, including metals and minerals, and studies their behaviors, like the dissociation of ionic compounds in solutions.
In the context of this exercise, we explored the inorganic complex \([\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{4}]\), involving metals like platinum and simple ions such as chloride. Investigating how these inorganic compounds interact in a solution provides insight into broader principles governing chemical reactions. Inorganic chemistry allows us to appreciate these dynamics and predict outcomes, like the necessity of 2 moles of \(\mathrm{AgNO}_3\) to react with dissociated chloride ions, forming \(\mathrm{AgCl}\).