Question

The solubility of \(\mathrm{AgCN}(\mathrm{s})\) in \(0.200 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) is \(8.8 \times 10^{-6} \mathrm{mol} / \mathrm{L} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{AgCN}\).

Short answer

The \(K_{sp}\) for \(AgCN\) is \(7.744 \times 10^{-11}\).

Step-by-step solution

Write the balanced chemical equation

First, we need to write out the balanced chemical equation. The chemical equation for the dissolution of \(AgCN\) in \(NH_3\) is: \[AgCN(s) + 2NH_3(aq) \leftrightarrow [Ag(NH_3)_2]^+ (aq) + CN^{-} (aq)\]

Write the expression for \(K_{sp}\)

Now, express the equilibrium constant for the reaction. In this case, it is called the solubility product constant (\(K_{sp}\)): \[K_{sp} = [[Ag(NH_3)_2]^+] \cdot [CN^-]\]

Substitution

As the solubility of \(AgCN\) is given to be \(8.8 \times 10^{-6}\) mol/L, this is the concentration of both [Ag(NH3)2]+ and CN- at equilibrium. Substitute this into our \(K_{sp}\) expression: \[K_{sp} = (8.8 \times 10^{-6}) \cdot (8.8 \times 10^{-6})\]

Calculate the value of \(K_{sp}\)

Finally, calculate the value of \(K_{sp}\) by multiplying these concentrations: \[K_{sp} = (8.8 \times 10^{-6})^2 = 7.744 \times 10^{-11}\]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry where the rate of the forward reaction equals the rate of the backward reaction.
This state of balance is crucial when studying reactions that occur in solution, especially in the formation of complex ions and solubility products.
When a reaction reaches equilibrium, the concentrations of the reactants and products remain constant because their rates of formation and consumption are equal.
This does not mean that the reactants and products are present in equal amounts, but that their concentrations do not change over time. Some key points about chemical equilibrium:

• The system is dynamic; reactions continue to occur, but no net change in concentration occurs.
• Examples include the dissolution and ionization processes where solid salts dissolve and reach equilibrium with their ions in solution.
• Equilibrium can be influenced by changes in concentration, temperature, and pressure.

Understanding chemical equilibrium is essential when considering how the solubility of substances like silver cyanide is affected in different environments, such as in the presence of ammonia.

Silver Cyanide Solubility

Silver cyanide ( AgCN) is a poorly soluble compound in water. However, its solubility can be improved in the presence of ammonia ( NH_3).
In the context of the provided exercise, we examine how AgCN dissolves in an aqueous ammonia solution to form a complex ion known as [Ag(NH_3)_2]^+. Silver cyanide's solubility is characterized by its solubility product constant, K_{sp}, which describes the equilibrium between the solid AgCN and its ions in solution.
This constant helps us understand how much of the compound will dissolve in a given solution. Some important notes about silver cyanide solubility:

• In pure water, AgCN has very limited solubility, leading to very low concentrations of Ag^+ and CN^- ions.
• The presence of ammonia increases the solubility significantly by forming the diammine silver complex, [Ag(NH_3)_2]^+.
• The formation of this complex affects the overall equilibrium and increases the solubility compared to when in pure water.

Knowing how silver cyanide behaves in various conditions helps predict its reaction behavior and practical applications.

Complex Ion Formation

The concept of complex ion formation is integral when studying the solubility and reactions of certain compounds in solvents like ammonia.
Complex ions are formed when a metal ion bonds with molecules or ions, which are called ligands. In the case of silver cyanide ( AgCN) dissolved in ammonia ( NH_3), a complex ion, [Ag(NH_3)_2]^+, forms.
This process greatly influences the solubility and chemical behavior of the compound. Here's how complex ion formation impacts solubility:

• The formation of [Ag(NH_3)_2]^+ reduces the concentration of Ag^+, shifting the equilibrium to dissolve more AgCN.
• Complex ion formation often results in increased solubility of sparingly soluble salts in strong ligand solutions.
• This phenomenon is useful in various extraction and purification processes in chemistry.

Understanding complex ion formation is essential, particularly in calculating the solubility product constant and predicting the behavior of solutions in chemical processes.