Problem 91 - Chemistry Matter and Change

Question

What is the solubility in mol/L of silver iodide, AgI. \(K_{\mathrm{sp}}\) for AgI is \(3.5 \times 10^{-17} .\) (Chapter 17\()\)

Step-by-Step Solution

Answer

The solubility of silver iodide (AgI) in mol/L is approximately \(5.92 \times 10^{-9}\) mol/L.

Step 1: Write the balanced dissolution reaction for AgI

First, write the balanced dissolution reaction for AgI: AgI(s) <=> Ag+ (aq) + I- (aq) This indicates that when silver iodide dissolves, it dissociates into silver (Ag+) and iodide (I-) ions in aqueous solution.

Step 2: Write the expression for Ksp

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. For the AgI dissolution reaction, the Ksp expression is: Ksp = [Ag+] [I-] where [Ag+] and [I-] represent the equilibrium concentrations of the silver and iodide ions, respectively.

Step 3: Define the solubility in terms of x

To find the solubility of AgI, let's call it x. When AgI dissociates, it produces equal amounts of Ag+ and I- ions. So, the concentration of Ag+ and I- ions can be represented by x: [Ag+] = x [I-] = x

Step 4: Substitute x into the Ksp expression and solve for x

Now, substitute the concentrations of Ag+ and I- into the Ksp expression and solve for x: Ksp = (x)(x) = x^2 Given Ksp is 3.5 x 10^-17, we have: x^2 = 3.5 x 10^-17 To solve for x, take the square root of both sides of the equation: x = sqrt(3.5 x 10^-17)

Step 5: Calculate the solubility of AgI

Using a calculator, solve for x: x = sqrt(3.5 x 10^-17) ≈ 5.92 x 10^-9 Therefore, the solubility of silver iodide (AgI) in mol/L is approximately 5.92 x 10^-9 mol/L.

Key Concepts

Ksp (solubility product constant)

The solubility product constant, abbreviated as \(K_{sp}\), is a unique constant that indicates the extent to which a sparingly soluble ionic compound can dissolve in water.

• It is an equilibrium constant specific to the dissolution process of ionic compounds.
• Its value is derived from the concentrations of the ions that form when the compound dissolves.

For instance, in the case of silver iodide (AgI), the \(K_{sp}\) tells us how much of the compound can dissolve before reaching saturation.
Since \(K_{sp}\) values are usually very small, this highlights the low solubility of these compounds, indicating that very few ions dissolve in water before coming to an equilibrium state.
Understanding \(K_{sp}\) is crucial in predicting whether a precipitate will form under different conditions.

Silver Iodide (AgI)

Silver iodide, represented as AgI, is an interesting compound because it has very low solubility in water.
Upon dissolution, it breaks into its components: silver ions (Ag⁺) and iodide ions (I⁻).
Despite its low solubility, it has important applications such as in photographic films and cloud seeding.

• AgI is used to create artificial precipitation, as its structure facilitates ice nucleation in clouds.
• The compound has a lattice structure that is seen in its solid form, but once it dissolves, it dissociates completely into its ionic form.

The dissolution reaction reflects this disassociation but keep in mind the complete dissolution is quite limited due to AgI's low solubility.

Dissolution Reaction

A dissolution reaction refers to the process by which a solid ionic compound separates into its constituent ions when introduced into a solvent, usually water.
This process is essential in understanding how substances like silver iodide (AgI) interact in solution.

• In the case of AgI, the dissolution can be represented as follows: AgI(s) \(\rightleftharpoons\) Ag⁺ (aq) + I⁻ (aq).
• It essentially describes how the solid (denoted "s") transitions into aqueous ions (denoted "aq").

Dissolution reactions are fundamental for determining the solubility of compounds and for calculating various equilibrium properties like \(K_{sp}\). Note that the reaction also signifies the reversible nature, where the product ions can potentially recombine into the solid form.

Equilibrium Concentrations

Equilibrium concentrations play a crucial role in understanding the dissolution process, particularly for sparingly soluble compounds.
For any dissolution reaction, achieving an equilibrium state means that the rate of dissolution equals the rate of reformation of the solid.

• When AgI dissolves, the concentrations of silver (Ag⁺) and iodide (I⁻) ions reach a point where further dissolution will not increase their concentrations.
• At this point, we use the variable "x" to represent these equilibrium concentrations since they are equal in the case of AgI.

Understanding equilibrium concentrations allows us to apply the \(K_{sp}\) expression, \(K_{sp} = [Ag⁺][I⁻]\), to determine the solubility of the compound.
Calculating the "x" variable helps find practical solubility in terms of mol/L. This highlights how equilibria in solutions determine possible reactions and concentrations.