Question

A 1.00-L solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

Short answer

The solubility-product constant (Ksp) for lead(II) iodide (PbI₂) at 25°C is calculated by first finding the molar concentration of Pb²⁺ and I⁻ ions in the saturated solution. In this case, we have [Pb²⁺] = 0.00117 M and [I⁻] = 0.00234 M. Using the Ksp expression, Ksp = [Pb²⁺][I⁻]², we find that Ksp = \(6.43 * 10^{-9} M^3\).

Step-by-step solution

Determine the number of moles of PbI₂ in 1.00 L of solution

First, we will convert the given mass of PbI₂ (0.54 g) into moles by using the molar mass of PbI₂. To find the molar mass, add the atomic mass of lead and twice the atomic mass of iodine: PbI₂ molar mass = 207.2 g/mol (lead) + 2 * 126.9 g/mol (iodine) = 460.98 g/mol Now, divide the mass of PbI₂ in the solution by its molar mass: Moles of PbI₂ = (0.54 g) / (460.98 g/mol) = 0.00117 mol Since we have 1.00 L of solution, the molar concentration of PbI₂ is: [PbI₂] = (0.00117 mol) / (1.00 L) = 0.00117 M

Write the chemical equation for the dissolution of PbI₂

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

Determine the molar concentration of Pb²⁺ and I⁻ ions

Since every mole of PbI₂ that dissolves produces one mole of Pb²⁺ ions and two moles of I⁻ ions, we have: [Pb²⁺] = 0.00117 M (from the dissociation of PbI₂) [2I⁻] = 2 * 0.00117 M = 0.00234 M

Calculate the solubility-product constant (Ksp) for PbI₂

We can now calculate the Ksp for PbI₂ using the Ksp expression: Ksp = [Pb²⁺] [I⁻]² Substitute the molar concentrations of Pb²⁺ and I⁻ ions: Ksp = (0.00117 M) * (0.00234 M)² = 6.43 * 10⁻⁹ M³ The solubility-product constant for lead(II) iodide at 25°C is 6.43 * 10⁻⁹ M³.

Key concepts

Lead(II) Iodide

Lead(II) iodide, represented by the chemical formula \( \text{PbI}_2 \), is a bright yellow solid that is sparingly soluble in water. It is a salt that consists of lead and iodide ions. When it dissolves in water, it separates into one lead ion and two iodide ions for each formula unit. This limited solubility is due to the strong ionic bonds between the lead and iodide ions, making it difficult for water to break them apart and dissolve the compound. This compound is often used in chemistry to study solubility equilibria and is known for producing vibrant yellow precipitates.

Understanding the properties of lead(II) iodide is crucial when performing solubility calculations and determining the solubility product constant \( K_{sp} \). Whether you're in the lab or solving equations, appreciating these properties can simplify many chemical equilibrium calculations.

Chemical Equilibrium

In the context of dissolving lead(II) iodide, chemical equilibrium plays an essential role. Chemical equilibrium occurs when the rate of the forward reaction (dissolution of \( \text{PbI}_2 \)) equals the rate of the reverse reaction (precipitation of \( \text{PbI}_2 \)). At this point, the concentration of reactants and products remains constant over time, entering a dynamic balance.

In the dissolution of lead(II) iodide, the chemical equation is \( \text{PbI}_2(s) \, \rightleftharpoons \, \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \). When equilibrium is reached, the rate at which \( \text{PbI}_2 \) dissolves in the solution is equal to the rate at which it is formed by the recombination of lead \( \text{Pb}^{2+} \) and iodide \( \text{I}^- \) ions. This balance allows us to determine the concentrations of the ions at equilibrium, which are used to calculate the solubility product constant \( K_{sp} \).

Molar Concentration

Molar concentration, often represented by the letter \([ ]\), is the amount of a substance (in moles) present in a unit volume of solution (usually in liters). It is an essential concept in solubility calculations and is crucial for understanding the behavior of ions in solution.

In the case of lead(II) iodide, the molar concentration of \( \text{PbI}_2 \) dissolved in solution can be determined directly from its known amount in grams and its molar mass. The resulting concentration is then used to find the concentrations of the ions produced when the solute dissolves.

For example:

• The molar mass of \( \text{PbI}_2 \) is 460.98 g/mol.
• The mass of \( \text{PbI}_2 \) in solution is 0.54 g.
• Using these, we calculate the moles: \( 0.54 \text{ g} / 460.98 \text{ g/mol} = 0.00117 \text{ mol} \).
• Thus, in a 1.00 L solution, the molar concentration of \( \text{PbI}_2 \) is 0.00117 M.

Understanding molar concentration is essential for calculating the extent to which a solute dissolves and for predicting how it will react in different chemical processes.

Solubility Calculation

Solving solubility problems often involves calculating the solubility product constant \( K_{sp} \). The \( K_{sp} \) is a specific type of equilibrium constant applied to the dissolution of sparingly soluble salts. For lead(II) iodide, knowing \( K_{sp} \) helps us quantify its solubility in water and predict how much of it will dissolve under equilibrium conditions.

The expression for \( K_{sp} \) of \( \text{PbI}_2 \) is derived from the equilibrium condition set by dissolving \( \text{PbI}_2 \): \( K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \). From earlier calculations, we found:

• \([\text{Pb}^{2+}] = 0.00117 \, \text{M} \)
• \([\text{I}^-] = 2 \, \times\, 0.00117 \, \text{M} = 0.00234 \, \text{M} \)

By substituting these values in, we calculate:
\[ K_{sp} = (0.00117) \times (0.00234)^2 = 6.43 \times 10^{-9} \text{ M}^3 \]

This calculation demonstrates how to derive \( K_{sp} \) from measured solubility data, offering insights into the solubility characteristics of difficult-to-dissolve compounds like lead(II) iodide. Knowing \( K_{sp} \) enables chemists to predict precipitation conditions and understand ionic equilibria in various chemical systems.