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\(\mathrm{Ag}^{+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)^{+}\right] ; \mathrm{K}_{1}=3.5 \times 10^{-3}\) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)\right]^{+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} ; \mathrm{K}_{2}=1.7 \times\) \(10^{-3}\) Then the formation constant of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is: (a) \(6.08 \times 10^{-6}\) (b) \(6.08 \times 10^{6}\) (c) \(6.08 \times 10^{-9}\) (d) None

Short Answer

Expert verified
The formation constant is \(5.95 \times 10^{-6}\), which matches option (d) None.

Step by step solution

01

Understand the Reaction

The formation of \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\) is a two-step process. In the first step, \(\mathrm{Ag}^+\) combines with \(\mathrm{NH}_3\) to form \([\mathrm{Ag}(\mathrm{NH}_3)]^+\), represented by the equilibrium constant \(K_1\). In the second step, \([\mathrm{Ag}(\mathrm{NH}_3)]^+\) reacts with another \(\mathrm{NH}_3\) to form \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\), with equilibrium constant \(K_2\). The overall formation constant is the product of these two equilibrium constants.
02

Calculate the Overall Formation Constant

To find the overall formation constant \(K_f\) for \([\mathrm{Ag}(\mathrm{NH}_3)_2]^+\), multiply the individual equilibrium constants: \[ K_f = K_1 \times K_2 \]Substitute the given values: \[ K_f = 3.5 \times 10^{-3} \times 1.7 \times 10^{-3} \]
03

Perform the Multiplication

Calculate the multiplication of the equilibrium constants: \[ K_f = (3.5 \times 1.7) \times (10^{-3} \times 10^{-3}) \]This yields: \[ K_f = 5.95 \times 10^{-6} \]
04

Compare with Given Options

Compare the calculated formation constant \(5.95 \times 10^{-6}\) with the options provided. The closest provided option is (d) None, since the computed value doesn't match any of the options exactly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In chemistry, the equilibrium constant, often denoted as \( K \), is an essential concept when discussing the balance of chemical reactions. It provides a measure of the ratio of the concentration of products to reactants at equilibrium. This constant is specific to a given reaction at a specific temperature.
A large equilibrium constant indicates that the reaction favors the formation of products, while a small constant suggests that reactants are favored.
For the formation of complex ions like the silver ammonia complex, equilibrium constants help us understand how readily the formation processes occur.
  • An equilibrium constant of \( K = 3.5 \times 10^{-3} \) shows that the formation is not highly favored.
  • It helps predict how much of the product will be formed at equilibrium.
The understanding of equilibrium constants involves balancing chemical equations, calculating concentrations, and predicting reaction behavior under various conditions. They are foundational in predicting the extent to which a reaction proceeds.
Complex Ion Formation
Complex ion formation is a fascinating process in coordination chemistry where metal ions react with ligands to form a structure known as a complex ion. This is central to understanding reactions like that of silver and ammonia to form silver ammonia complexes.
Key aspects to remember about complex ions:
  • They consist of a central metal ion bonded to surrounding ligands.
  • The formation often occurs in multiple steps, each described by its equilibrium constant.
  • These reactions are crucial in various industrial and biological processes, including catalysis and medication delivery.
Complex ion formation provides insights into the reactivity and interactions between metals and molecules. By analyzing these interactions, chemists can manipulate the reactions for desired outcomes, from synthesizing new compounds to removing pollutants. Understanding these processes is instrumental in mastering the field of coordination chemistry.
Silver Ammonia Complex
The silver ammonia complex is a common example in coordination chemistry where we see practical applications of complex ion formation. The process begins with the silver ion \( \mathrm{Ag}^+ \) reacting with ammonia \( \mathrm{NH}_3 \) to form intermediate and final products.
Highlights of the silver ammonia complex formation:
  • Initial interaction between \( \mathrm{Ag}^+ \) and \( \mathrm{NH}_3 \) forms \( [\mathrm{Ag}(\mathrm{NH}_3)]^+ \).
  • This intermediate further reacts with additional ammonia to eventually yield \( [\mathrm{Ag}(\mathrm{NH}_3)_2]^+ \).
  • These reactions are reversible and governed by their respective stepwise equilibrium constants \( K_1 \) and \( K_2 \).
This sequential formation highlights the intricacies involved in forming stable complex ions, essential for various applications such as photographic processes and in some analytical techniques. Learning about such complexes enhances our understanding of chemical reactivity and stability in solution.
Stepwise Formation Constants
Stepwise formation constants are key to understanding the formation of complex ions through sequential reactions, such as the creation of silver ammonia complexes. Each step in the formation has its associated equilibrium constant, which reflects the formation's ease or difficulty.
Here’s what you need to know about stepwise constants:
  • Each constant \( K_1, K_2, \) etc., corresponds to a specific step in the formation of a complex ion.
  • The overall formation constant \( K_f \) is the product of these stepwise constants.
  • In practical terms, the constants determine the conditions needed to achieve desired products or maintain certain concentrations of ions.
  • In our example of \( [\mathrm{Ag}(\mathrm{NH}_3)_2]^+ \), the stepwise constants are used extensively to calculate the total formation constant.
Understanding and calculating these constants is crucial in chemical research and industry, providing a basis for manipulating and predicting the behaviors of complex ions during reactions. They help chemists design and control processes in environmental clean-ups, medical therapies, and more.

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Most popular questions from this chapter

A reversible reaction is said to have attained equilibrium, when: (a) Backward reaction stops (b) Both backward and forward reactions take place at equal speed (c) Both backward and forward reactions stop (d) Concentration of each of the reactants and products becomes equal

The rate constants for the forward and backward reactions of hydrolysis of ester are \(1.1 \times 10^{-2}\) and \(1.5 \times\) \(10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. The equilibrium constant of the reaction, \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is: (a) \(6.53\) (b) \(7.34\) (c) \(7.75\) (d) \(8.33\)

At \(550 \mathrm{~K}\), the \(\mathrm{K}\) for the following reaction is \(10^{4} \mathrm{~mol}^{-1}\) lit \(\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})\) At equilibrium, it was observed that \([\mathrm{X}]=\frac{1}{2}[\mathrm{Y}]=\frac{1}{2}[\mathrm{Z}]\) What is the value of \([\mathrm{Z}]\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(2 \times 10^{-4}\) (b) \(10^{-4}\) (c) \(2 \times 10^{4}\) (d) \(10^{4}\)

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(K_{p}\) and \(K_{c}\) are related as: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}\) (c) \(K_{c}=K_{p} \times(R T)^{2}\) (d) \(K_{p}=K_{c} \times(R T)^{-2}\)

In which of the following reactions, equilibrium is independent of pressure: (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)

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