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Calculate the solubility in grams per liter of \(\operatorname{AgBr}(s)\) in \(0.35 \mathrm{M} \mathrm{NH}_{3}(a q) .\) Take \(K_{i}=2.0 \times 10^{7} \mathrm{M}^{-2}\) for the complexation reaction of \(\mathrm{Ag}^{+}(a q)\) to produce \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q)\). Assume a temperature of \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility substantially increases due to complexation, approaching equilibrium with a near-complete conversion into [Ag(NH鈧)鈧俔鈦.

Step by step solution

01

Understanding the Problem

We are asked to determine the solubility of silver bromide (AgBr) in a solution containing 0.35 M ammonia (NH鈧). We need to consider the formation of a coordination complex, namely silver diammine ([Ag(NH鈧)鈧俔鈦), facilitated by the coordination with NH鈧 with a given equilibrium constant.
02

Writing the Relevant Equations

The formation constant for the complexation reaction is: \[ \text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) \] with \( K_i = 2.0 \times 10^7 \text{ M}^{-2} \). Additionally, let's write the solubility equilibrium for silver bromide: \[ \text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq). \]
03

Analyzing the Solubility Product

The solubility product (Ksp) of AgBr is not provided but needs to be considered in combination with the formation of the complex ion. The ion concentrations manipulate the equilibrium in solution. Although we don't know directly the Ksp of AgBr, we'll proceed assuming it is significantly affected by the formation constant (as is common in complexing agent scenarios).
04

Consequences on Solubility

Since \(0.35 \text{ M}\) of NH鈧 can effectively drive the production of the complex due to its favorable equilibrium constant, the equilibrium of the AgBr solubility shifts predominately towards forming the complex. We can assume the concentration of Ag鈦 becoming negligible, leading to increased solubility.
05

Calculate Solubility In Terms of Initial Concentration

Given the equilibrium constant and high concentration of NH鈧:1. Determine the effective concentration of complexed Ag鈦, considering the entire solubility as transitioned to the form [Ag(NH鈧)鈧俔鈦.2. Assume nearly complete transition such that:\[ K_i = \frac{[\text{Ag}(\text{NH}_3)_2^+]}{[ ext{Ag}^+][\text{NH}_3]^2} = \text{high value}n\Rightarrow \frac{[\text{complex}]}{S[0.35]^2} \approx 2.0 \times 10^7 \text{ M}^{-2}.\]
06

Calculating Final Solubility Value

If phenotypically the denominator dominates concludes almost transition: Substitute known constants to solve for S, allowing the equilibrium to drive solubility into almost fully complexed form.

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Key Concepts

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

Understanding the Solubility Product
The concept of solubility product, denoted as \(K_{sp}\), is crucial in understanding the solubility of sparingly soluble salts. For example, in the given exercise, we must comprehend how the presence of ammonia affects the dissolution of silver bromide (AgBr). \[ \text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq) \] where the solubility product equilibrium is established for the dissociation of AgBr into its ions.
When a solid like AgBr is added to water, it dissolves into a dynamic equilibrium, balancing between the solid and its dissolved ions. The \(K_{sp}\) is the constant that describes this equilibrium and is calculated by multiplying the concentrations of the ions in solution at equilibrium. Essentially, a lower \(K_{sp}\) value indicates lower solubility.
In scenarios involving complex ion formation, such as with the introduction of an ammonia solution, the \(K_{sp}\) alone does not suffice to predict solubility and we must consider the effect of these complexes, which we'll discuss next.
The Role of Coordination Complexes
In the realm of chemistry, coordination complexes play a significant role, particularly in enhancing the solubility of compounds that would otherwise be sparingly soluble. At the heart of this topic is the interaction between a central metallic ion and multiple ligands like NH鈧.
In the given exercise, when ammonia (NH鈧) is present in solution, it acts as a ligand, binding to the silver ions (Ag鈦) and forming a coordination complex:
\[ \text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) \]
This reaction represents the formation of the diammine silver ion complex. As ammonia molecules are added, they stabilize the silver ion in solution, reducing its concentration as a free ion and driving the solubility of AgBr higher than it would be in pure water. This process is crucial for the complete transition of Ag鈦 ions to [Ag(NH鈧)鈧俔鈦, practically elevating the solubility of the original salt.
These coordination complexes hence play a profound role in altering the apparent solubility of certain salts through their ability to complex cations effectively.
Understanding Equilibrium Constants
Equilibrium constants are at the core of predicting the direction and extent of chemical reactions, especially involving complex ions. For the formation of a coordination complex, like the one we've discussed, the equilibrium constant is the formation constant, represented as \(K_i\).
In our scenario, \(K_i\) is given as \(2.0 \times 10^7 \, \text{M}^{-2}\), which is significantly large. This indicates a strong tendency for the Ag鈦 ions to form the complex ion [Ag(NH鈧)鈧俔鈦 when NH鈧 is available in ample concentration.
This constant emerges from the relationship:
\[ K_i = \frac{\left[\text{Ag}(\text{NH}_3)_2^+\right]}{[\text{Ag}^+][\text{NH}_3]^2} \]
The substantial magnitude of \(K_i\) in this context implies that the concentration of the complex ion at equilibrium is very high compared to the concentrations of the unbound silver and ammonia ions. This assures us that the equilibrium heavily favors the formation of the coordination complex, subsequently affecting the solubility of the original salt, AgBr.
Thus, understanding how equilibrium constants are derived and interpreted is essential to grasp complex formation and its effect on solubility dynamics.

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Most popular questions from this chapter

Calculate the solubility of \(\operatorname{CuS}(s)\) in a solution buffered at \(\mathrm{pH}=2.00\) and saturated with hydrogen sulfide so that \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.10 \mathrm{M}\). Assume a temperature of \(25^{\circ} \mathrm{C}\).

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