Problem 66 - Chemistry and Chemical Reactivity

Question

Solid silver iodide, AgI, can be dissolved by adding aqueous sodium cyanide to it. $$\mathrm{AgI}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftarrows\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})$$ Show that this equation is the sum of two other equations, one for dissolving AgI to give its ions and the other for the formation of the \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) ion from \(\mathrm{Ag}^{+}\) and \(\mathrm{CN}^{-}\). Calculate \(K_{\text {net }}\) for the overall reaction.

Step-by-Step Solution

Answer

Combine the dissolution and complex formation reactions; the net equilibrium constant is \( K_\text{net} = K_1 \times K_2 \).

Step 1: Write Individual Reactions

First, write the reaction for the dissolution of silver iodide (AgI) into its ions: \[ \mathrm{AgI} (s) \rightleftharpoons \mathrm{Ag}^+ (aq) + \mathrm{I}^- (aq). \] Next, write the reaction for the formation of the \([\mathrm{Ag}(\mathrm{CN})_{2}]^-\) complex ion: \[ \mathrm{Ag}^+ (aq) + 2 \mathrm{CN}^- (aq) \rightleftharpoons [\mathrm{Ag(CN)}_{2}]^- (aq). \]

Step 2: Add the Reactions

Combine these two reactions by adding them together. The silver ion \( \mathrm{Ag}^+ \) is produced in the first reaction and consumed in the second, so it cancels out, resulting in the overall reaction: \[ \mathrm{AgI} (s) + 2 \mathrm{CN}^- (aq) \rightleftharpoons [\mathrm{Ag(CN)}_{2}]^- (aq) + \mathrm{I}^- (aq). \]

Step 3: Identify Equilibrium Constants

Let \( K_1 \) be the equilibrium constant for the dissolution of AgI: \[ K_1 = K_{sp} = [\mathrm{Ag}^+] [\mathrm{I}^-]. \] For the formation of \([\mathrm{Ag}(\mathrm{CN})_{2}]^-\), let the equilibrium constant be \( K_2 \): \[ K_2 = \frac{[[\mathrm{Ag(CN)}_{2}]^-]}{[\mathrm{Ag}^+][\mathrm{CN}^-]^2}. \]

Step 4: Calculate Overall Equilibrium Constant \( K_\text{net} \)

The equilibrium constant for the overall reaction, \( K_\text{net} \), is the product of \( K_1 \) and \( K_2 \): \[ K_\text{net} = K_1 \times K_2. \] This means, \[ K_\text{net} = \frac{[[\mathrm{Ag(CN)}_{2}]^-][\mathrm{I}^-]}{[\mathrm{CN}^-]^2}. \]

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial part of understanding chemical equilibrium. It helps us quantify the balance between the reactants and products in a reversible reaction. In the case of the dissolution of silver iodide (AgI) and complex ion formation, two separate reactions are considered.

The first reaction is the dissolution of AgI into its ions:

• \( \text{AgI} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \).

The equilibrium constant for this process is the solubility product \( K_{sp} \), which is expressed as:

• \( K_1 = [\text{Ag}^+][\text{I}^-] \).

The second reaction involves the formation of a complex ion:

• \( \text{Ag}^+ (aq) + 2 \text{CN}^- (aq) \rightleftharpoons [\text{Ag(CN)}_{2}]^- (aq) \).

For this, the equilibrium constant \( K_2 \) is:

• \( K_2 = \frac{[[\text{Ag(CN)}_{2}]^-]}{[\text{Ag}^+][\text{CN}^-]^2} \).

The overall equilibrium constant \( K_{net} \) for the combined reaction is the product of these two constants:

• \( K_{net} = K_1 \times K_2 \).

This calculation provides insight into how the reaction progresses under equilibrium conditions.

Complex Ion Formation

Complex ion formation is an essential part of the process where a metal cation bonds with ligands, forming a complex. In our scenario, the complex ion \([\text{Ag(CN)}_{2}]^-\) is formed when silver ions \((\text{Ag}^+)\) react with cyanide ions \((\text{CN}^-)\).

This is represented by the reaction:

• \( \text{Ag}^+ (aq) + 2 \text{CN}^- (aq) \rightleftharpoons [\text{Ag(CN)}_{2}]^- (aq) \).

Formation of these complex ions is influenced by the concentration of the ions in the solution, which significantly shifts the equilibrium.

Practically, the creation of a complex ion can increase the solubility of a salt. This concept is crucial in understanding how substances like AgI, which are generally insoluble, can be dissolved by complexation.

The ability to form such complex ions is determined by factors like the ligand's affinity for the metal ion, which is identifiable by the value of \( K_2 \), the formation constant. A larger \( K_2 \) value signifies a more stable complex.

Silver Iodide Dissolution

Silver iodide (AgI) is known for its low solubility in water. However, the addition of sodium cyanide allows for its dissolution through complex ion formation. The initial reaction dissolves AgI into silver ions \((\text{Ag}^+)\) and iodide ions \((\text{I}^-\)).

This reaction can be described by:

• \( \text{AgI} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \).

The solubility product \( K_{sp} \) explains how much AgI can dissolve in water. It is inherently small for silver iodide, indicating its low solubility. However, the introduction of cyanide ions \((\text{CN}^-)\) leads to the accumulation of \([\text{Ag(CN)}_{2}]^-\), shifting the equilibrium significantly.

This process illustrates the principle of Le Chatelier's, where the reaction adjusts itself to counterbalance any changes, such as the addition of cyanide, ultimately enhancing the solubility of AgI.

Solute-Solvent Interactions

Solute-solvent interactions determine how a substance dissolves in a solvent. In the context of silver iodide's dissolution with sodium cyanide, these interactions are crucial.

Here, sodium cyanide acts as a source of cyanide ions \((\text{CN}^-)\) which facilitates the formation of soluble complexes. The dissolution of AgI itself involves breaking apart the solid into its ions, demanding energy to overcome lattice forces.

Cyanide ions then interact with silver ions to form \([\text{Ag(CN)}_{2}]^-\). This complex's soluble nature stems from strong interactions between \( \text{Ag}^+ \) and \( \text{CN}^- \).

These interactions are a vivid display of solubility principles and illustrate why certain otherwise insoluble substances can dissolve under specific conditions. Factors like ion size, charge, and the ability to form new interactions contribute to understanding how solute-solvent dynamics work.