Problem 56 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

The solubility-product constant, \( K_{sp} \), for lead(II) iodide at \(25^{\circ}\)C is approximately \(6.42 \times 10^{-9}\).

Step 1: Understanding the Dissolution Equation

The dissolution of lead(II) iodide in water can be represented by the equation: \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]This means that when lead(II) iodide dissolves, it produces one lead ion and two iodide ions.

Step 2: Calculate the Moles of Lead(II) Iodide

We begin by calculating the molar mass of \( \text{PbI}_2 \). The atomic masses are approximately: \( \text{Pb} = 207.2 \) and \( \text{I} = 126.9 \). Thus, \[ \text{Molar mass of PbI}_2 = 207.2 + 2(126.9) = 461 \text{ g/mol} \]Now, calculate the moles of \( \text{PbI}_2 \) in the solution:\[ \text{Moles of PbI}_2 = \frac{0.54 \text{ g}}{461 \text{ g/mol}} \approx 0.00117 \text{ mol} \]

Step 3: Determine Ion Concentrations

From the dissolution equation, each mole of \( \text{PbI}_2 \) dissolves into 1 mole of \( \text{Pb}^{2+} \) and 2 moles of \( \text{I}^- \). Therefore:- \( \text{[Pb}^{2+}] = 0.00117 \text{ mol/L}\)- \( \text{[I}^-] = 2 \times 0.00117 \text{ mol/L} = 0.00234 \text{ mol/L}\)

Step 4: Calculate the Solubility-Product Constant

The solubility-product constant \( K_{sp} \) for \( \text{PbI}_2 \) is given by:\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]Substitute the ion concentrations:\[ K_{sp} = (0.00117)(0.00234)^2 \]Performing the calculation:\[ K_{sp} = 0.00117 \times 0.00234 \times 0.00234 \approx 6.42 \times 10^{-9} \]

Key Concepts

Solubility-Product Constant (Ksp)

The solubility-product constant, often abbreviated as Ksp, is a vital concept in chemistry, particularly when studying solubility equilibria. It helps us understand the extent to which a compound can dissolve in water. The Ksp is an equilibrium constant that applies to a sparingly soluble ionic compound. This means that for a dissolved ionic compound in a saturated solution, the multiplication of the concentrations of the ions each raised to the power of their coefficients in the balanced dissolution equation remains constant at a given temperature. For example, when dealing with lead(II) iodide, PbI₂, the Ksp reflects the product of the molarity of lead ions and the square of the molarity of iodide ions in the solution. The Ksp can help predict how much of the compound will dissolve in the water under standard conditions. It’s crucial for anticipating the behavior of ions in diverse chemical environments, from laboratory settings to environmental systems.

Dissolution Equation

In chemistry, the dissolution equation provides a clear picture of what happens when a solid solute dissolves in a solvent, typically water. For lead(II) iodide, the dissolution reaction can be written as: \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \] This equation shows that one molecule of lead(II) iodide, a solid, separates into one lead ion and two iodide ions when it enters the solution. Understanding this equation is crucial because it tells us the stoichiometric relationships between the reactants and the products. This dissolution process dynamically reaches equilibrium, meaning while some of the solid dissolves into ions, some ions will recombine into the solid, maintaining a steady state at equilibrium. This equilibrium is what allows us to use Ksp to determine the solubility characteristics of compounds.

Ion Concentrations in Solution

Calculating ion concentrations in a solution is a step that follows the dissolution equation. The process begins after determining the number of moles of the solute present. For lead(II) iodide, after finding the moles, you can deduce the concentrations of dissolved ions. One mole of PbI₂ yields one mole of \( \text{Pb}^{2+} \) ions and two moles of \( \text{I}^- \) ions. Therefore, if you have 0.00117 moles of PbI₂ dissolved in a liter of solution, you have 0.00117 M of \( \text{Pb}^{2+} \) and 0.00234 M of \( \text{I}^- \). This calculation is essential for finding Ksp because ion concentrations precisely determine the saturation point of the compound in the solution. These quantitative insights provide valuable information on which conditions could lead to precipitation or continued dissolution, aiding in both academic and practical chemical problem-solving.