Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given : \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+} \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{NH}_{3}, K_{c}=6.2 \times 10^{-8}\) and \(K_{s p}\) of \(\mathrm{AgCl}=1.8 \times 10^{-10}\) at \(298 \mathrm{~K}\). If ammonia is added to a water solution containing excess of \(\mathrm{AgCl}(\mathrm{s})\) only, calculate the concentration of the complex in \(1.0 \mathrm{M}\) aqueous ammonia.

Step-by-Step Solution

Answer

The concentration of the complex is approximately 2.9 mM.

Step 1: Write the Relevant Equilibrium Expressions

For the complex dissociation equilibrium \[\mathrm{Ag(NH}_3)_2^+ \rightleftharpoons \mathrm{Ag}^+ + 2 \mathrm{NH}_3, \]the equilibrium constant is \[ K_c = \frac{[\mathrm{Ag}^+][\mathrm{NH}_3]^2}{[\mathrm{Ag(NH}_3)_2^+]} = 6.2 \times 10^{-8}. \]

Step 2: Express Solubility of AgCl

For the solubility equilibrium of silver chloride:\[\mathrm{AgCl}_{(s)} \rightleftharpoons \mathrm{Ag}^+ + \mathrm{Cl}^-, \]the solubility product is \[ K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] = 1.8 \times 10^{-10}. \]

Step 3: Consider Concentration of Ammonia

Assume that the concentration of ammonia in the solution remains approximately 1 M due to the presence of excess ammonia.

Step 4: Avoid Complete Precipitation

Given that ammonia forms a complex with silver ions, the presence of excess \(\mathrm{NH}_3\) pushes the equilibrium:\[\mathrm{Ag(NH}_3)_2^+ \rightleftharpoons \mathrm{Ag}^+ + 2 \mathrm{NH}_3\]to the left, ensuring silver ions are present in solution.

Step 5: Calculate Equilibrium Concentration of Complex

Let the concentration of \(\mathrm{Ag(NH}_3)_2^+\) be \(x\). Then, \[\mathrm{Ag}^+ = \frac{x}{K_c} \left(\frac{1}{[\mathrm{NH}_3]^2}\right) = x \cdot 6.2 \times 10^{-8} \times \frac{1}{1^2} \approx 6.2 \times 10^{-8}x.\]

Step 6: Solve for x Using Ksp

Plug \( [\mathrm{Ag}^+] \approx 6.2 \times 10^{-8}x \) into the solubility expression:\[ 1.8 \times 10^{-10} = (6.2 \times 10^{-8} \times x)[\mathrm{Cl}^-] \]With \([\mathrm{Cl}^-] = 1.0 \textrm{ M}\) due to saturation:\[ 1.8 \times 10^{-10} = 6.2 \times 10^{-8}x \times 1 \]Solve for \(x\):\[ x \approx \frac{1.8 \times 10^{-10}}{6.2 \times 10^{-8}} \approx 2.9 \times 10^{-3} \text{ M}.\]

Step 7: Conclude with Final Concentration

The concentration of \(\mathrm{Ag(NH}_3)_2^+\) in the solution is approximately 2.9 mM.

Key Concepts

Complex Ion Formation

When certain metals, such as silver, interact with various ligands, they form complex ions. In our exercise, silver ions (\( \text{Ag}^+\) ) in solution can interact with ammonia (\( \text{NH}_3\) ). When these components combine, they form a complex ion: \( \text{Ag(NH}_3)_2^+\). This occurs because ammonia acts as a ligand, donating its lone pair of electrons to the silver ion, thereby stabilizing it in the solution.
Complex ions are essential in chemistry because they significantly alter the behavior of ions in solution, affecting properties such as solubility and reactivity. Understanding how they form and behave can help predict the outcomes of various chemical reactions.

• The formation process involves ligand-to-metal binding.
• Complex ion formation can greatly change the chemical equilibrium of a system.

In this exercise, the presence of ammonia pushes the reaction towards forming the complex ion, which affects the solubility of silver chloride.

Solubility Product

The solubility product constant, often abbreviated as \( K_{sp} \), is a crucial concept when dealing with slightly soluble ionic compounds. It represents the product of the concentrations of the ions involved, each raised to the power of their stoichiometric coefficients, at equilibrium. For our exercise, the dissolution of silver chloride (\( \text{AgCl} \)) is expressed as:
\[ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
The solubility product for this equilibrium is given as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.8 \times 10^{-10} \]
Here are some key points to remember:

• \( K_{sp} \) is specific for each salt at a given temperature.
• It provides insight into how much of the salt can dissolve in solution before the start of precipitation.

In the context of our problem, understanding \( K_{sp} \) of silver chloride helps us determine the concentration of ions before and after complex formation with ammonia.

Equilibrium Constant

Equilibrium constants are vital in predicting the ratio of concentrations of products to reactants at equilibrium in a chemical reaction. Specifically, the equilibrium constant \( K_c \) for the complex dissociation of \( \text{Ag(NH}_3)_2^+ \) provides insight into the formation and dissociation of this complex ion in solution:
\[ \text{Ag(NH}_3)_2^+ \rightleftharpoons \text{Ag}^+ + 2\text{NH}_3 \]
\[ K_c = \frac{[\text{Ag}^+][\text{NH}_3]^2}{[\text{Ag(NH}_3)_2^+]} = 6.2 \times 10^{-8} \]
Equilibrium constants help us determine the direction of the reaction under certain conditions, especially:

• If \( K_c \) is large, the reaction favors products.
• If \( K_c \) is small, as in our exercise, the reaction favors reactants.

In this case, the presence of excess ammonia further drives the reaction towards maintaining the complex ion, reducing free \( \text{Ag}^+ \) ions.

Concentration Calculations

Concentration calculations are the backbone of quantitative chemistry in equilibrium systems. In our scenario, we needed to compute the concentration of \( \text{Ag(NH}_3)_2^+ \) in solution. Begin with the known values and equilibrium conditions:

• The ammonia concentration remains approximately 1 M due to excess.
• Use the value of \( K_c \) to relate the concentrations of ions at equilibrium.
• Solve for \( x \), where \( x \) is the equilibrium concentration of \( \text{Ag(NH}_3)_2^+ \).

By substituting \( [\text{Ag}^+] = 6.2 \times 10^{-8}x \) into the solubility product expression, the final equilibrium concentration of the complex is calculated as \( 2.9 \text{ mM} \). This exercise highlights how to handle equilibrium calculations to find concentrations.
It illustrates using equilibrium expressions to deduce unknown concentrations from known constants.

Ammonia Complexation

Ammonia is a common ligand used in forming complex ions due to its electron-rich nature. In this exercise, ammonia's ability to form a complex with silver ions underlies the central concept of complexation:
\[ \text{Ag}^+ + 2\text{NH}_3 \rightleftharpoons \text{Ag(NH}_3)_2^+ \]
Ammonia provides electron lone pairs to the silver ion, stabilizing it in a less reactive complex form and affecting the entire equilibrium system. Key aspects of ammonia complexation include:

• Ligand interactions heavily influence metal ion solubility.
• The formation of such complexes often shifts the equilibrium, affecting overall reactivity.

In solutions with ammonia, the likelihood of forming complexes like \( \text{Ag(NH}_3)_2^+ \) increases, reducing the concentration of free metal ions. This process is fundamental in applications ranging from analytical chemistry to industrial processes, where controlling ion behavior is crucial.