Problem 32 - Chemistry in 30 days

Question

The solubility product of \(\mathrm{Agl}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16}\) \(\mathrm{mol}^{2} \mathrm{~L}^{-2}\). The solubility of Agl in \(10^{-4} \mathrm{~N}\) solution of \(\mathrm{KI}\) at \(25^{\mathrm{"}} \mathrm{C}\) is approximately (in \(\left.\operatorname{mol} \mathrm{L}^{-1}\right)\) (a) \(1.0 \times 10^{-16}\) (b) \(1.0 \times 10^{-12}\) (c) \(1.0 \times 10^{-10}\) (b) \(1.0 \times 10^{-12}\) (d) \(1.0 \times 10^{-8}\)

Step-by-Step Solution

Answer

The solubility of AgI in \(10^{-4} \text{ N} \text{ KI}\) is approximately \(1.0 \times 10^{-12} \text{ mol L}^{-1}\), so the answer is (b).

Step 1: Write the Dissociation Reaction

The compound silver iodide (AgI) dissociates in water according to the following equilibrium reaction: \[ \text{AgI} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \].

Step 2: Setup the Expression for Solubility Product (Ksp)

The solubility product expression for AgI is \[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]. The given solubility product (Ksp) is \(1.0 \times 10^{-16}\).

Step 3: Consider the Common Ion Effect

Since the solution contains 0.0001 M KI, it will completely dissociate to give \([\text{I}^-] = 1.0 \times 10^{-4} \text{ M}\), which affects the solubility of AgI due to the common ion effect.

Step 4: Solve for the Concentration of Ag+

With the common ion effect, the new solubility equilibrium expression becomes \[ K_{sp} = [\text{Ag}^+](1.0 \times 10^{-4} + x) \approx [\text{Ag}^+] \times 1.0 \times 10^{-4} \]. Thus, \([\text{Ag}^+] = \frac{1.0 \times 10^{-16}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-12} \text{ M}\).

Step 5: Approximate the Solubility of AgI

The concentration of \(\text{Ag}^+\) represents the solubility of AgI under the influence of the common ion effect. Therefore, the solubility of AgI in \(10^{-4} \text{ N} \text{ KI}\) is approximately \(1.0 \times 10^{-12} \text{ mol L}^{-1}\).

Key Concepts

Common Ion Effect

The common ion effect is a phenomenon that occurs when an electrolyte that shares a common ion with a dissolved substance is added to the solution. This can significantly impact the solubility of the dissolved compound. In simpler terms, if the "common ion" is already present in the solution, it can reduce the dissolution of the ionic compound.

In our silver iodide (AgI) exercise, adding potassium iodide (KI) introduces the iodide ion \(I^-\) into the solution, which is already a component of the dissociated AgI. This increase in \(I^-\) concentration pushes the dissociation equilibrium of AgI back towards the solid form, thus reducing its solubility.

Here's how the common ion effect works:

• The presence of additional iodide ions from KI reduces the concentration of silver ions \(\left(\text{Ag}^+\right)\).
• This happens because the solution reaches equilibrium faster, decreasing the solubility of AgI as per Le Chatelier's principle.

Overall, the common ion effect is a key factor in determining how dissolved ions affect a solution's equilibrium.

Silver Iodide

Silver iodide is an intriguing compound with unique properties. It's represented by the chemical formula AgI and is known for its low solubility in water. When dissolved, it dissociates into silver ions \(\text{Ag}^+\) and iodide ions \(\text{I}^-\). However, this dissociation is limited due to its insolubility.

Why is silver iodide important in chemistry and everyday applications? Let's explore:

• Weather modification: Silver iodide is used in cloud seeding to encourage precipitation. Its crystalline structure can initiate ice crystal formation, enhancing rainfall.
• Photography: In the past, AgI was used in photographic emulsions. When exposed to light, it forms elemental silver, creating an image.
• Study of solubility equilibria: AgI serves as a classic example for explaining and understanding solubility equilibria, especially with the presence of common ions.

In chemical reactions and experimental conditions, AgI's balance between its low solubility and diverse applications highlights its vital role.

Equilibrium Reactions

Equilibrium reactions describe the state in which the forward and reverse reactions occur at the same rate. In this balanced state, the concentration of reactants and products remains constant over time.

For silver iodide, the equilibrium reaction is:\[\text{AgI} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq)\]Here, solid AgI dissociates into its ions, with solubility product \(K_{sp}\) expressing the equilibrium condition.

Key components of equilibrium reactions include:

• Forward and reverse rates: Both reactions proceed at equal rates at equilibrium.
• Static concentrations: Once equilibrium is achieved, concentrations don't change unless disturbed.
• Solubility product (\(K_{sp}\)): It's a constant for sparingly soluble compounds, indicating the product of ionic concentrations.

In practice, disturbances such as the previous introduction of a common ion (from the KI solution) will shift this equilibrium, as per Le Chatelier's principle. This shift changes the concentration of ions in the solution and illustrates the dynamic nature of equilibrium. Understanding equilibrium reactions helps predict how substances will behave under different conditions.