Question

A \(1.00-\mathrm{L}\) 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 PbI2 at 25°C is \( 6.41 \times 10^{-9} \).

Step-by-step solution

Write the balanced dissociation equation for PbI2

The balanced dissociation equation for PbI2 in water is: \[ \mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)} \]

Write the expression for the solubility-product constant, Ksp

Ksp is the product of the equilibrium concentrations of the ions in the saturated solution, each raised to the power of their stoichiometric coefficients. In this case, the expression becomes: \[ \mathrm{K_{sp} = [Pb^{2+}][I^-]^2} \]

Calculate the molar concentrations of the ions in the saturated solution

Given that the mass of dissolved PbI2 is 0.54 g in 1 L of the saturated solution, we can find the moles of \(\mathrm{PbI}_2\) by dividing the mass by its molar mass: \[ \frac{0.54 \,\mathrm{g}}{461.01 \,\mathrm{g/mol}} = 0.00117 \,\mathrm{mol} \] Since 1 mol of PbI2 dissociates into 1 mol of Pb²⁺ and 2 mol of I⁻ ions, we can calculate the molar concentration of each ion in the saturated solution: \[ [\mathrm{Pb^{2+}}] = 0.00117 \,\mathrm{M} \] \[ [\mathrm{I^-}] = 2 \times 0.00117 \,\mathrm{M} = 0.00234 \,\mathrm{M} \]

Substitute the molar concentrations into the Ksp expression and solve for Ksp

Now, we can substitute the calculated molar concentrations of Pb²⁺ and I⁻ ions into the Ksp expression: \[ \mathrm{K_{sp} = (0.00117)(0.00234)^2 =}\, 6.41 \times 10^{-9} \] The solubility-product constant for PbI2 at 25°C is \( 6.41 \times 10^{-9} \).

Key concepts

Lead(II) Iodide

Lead(II) Iodide, denoted as PbI₂, is a chemical compound that prominently features in discussions about solubility because of its sparingly soluble nature. It is known for its striking bright yellow color, which can serve as an excellent visual indicator in laboratory settings. The compound is composed of one lead (Pb) ion paired with two iodide (I⁻) ions. At room temperature, Lead(II) Iodide exists in a solid state when it is not dissolved in a solvent.

To understand PbI₂ more thoroughly, one must consider its molecular structure and the interactions between ions. The lead ion (Pb²⁺) forms a stable coordination with the iodide ions, which is why under normal conditions, PbI₂ remains largely undissolved. However, when dissolved in water, it dissociates into its constituent ions. The degree to which this dissociation happens is governed by its Solubility Product Constant (Ksp). Understanding this constant helps predict the solubility of PbI₂ when environmental conditions change. Such insights are important in various fields, ranging from environmental science to industrial applications.

Chemical Equilibrium

Chemical equilibrium refers to a state in which the forward and backward reactions occur at the same rates, leading to constant concentrations of reactants and products. When applied to the case of Lead(II) Iodide, chemical equilibrium is achieved in a saturated solution where the rate of dissolution of PbI₂ equals the rate of its precipitation.

In a chemical context, this balance is expressed through the equilibrium constant expression, in this case, the solubility product constant or Ksp. For lead(II) iodide in solution, the dissociation can be represented as:

• PbI₂ (s) ⇌ Pb²⁺ (aq) + 2 I⁻ (aq)

The solubility product expression is derived from this equation and is pivotal in determining at what concentration a compound reaches equilibrium. This expression includes the concentrations of the ions raised to the power of their coefficients in the balanced equation. By knowing the Ksp, chemists can calculate the concentrations of dissolved ions at equilibrium, which in turn helps predict potential precipitate formation or dissolution upon changing conditions.

Dissolution Reaction

A dissolution reaction is a process where a solid substance dissolves in a solvent, forming a solution. In the case of Lead(II) Iodide, the dissolution reaction is represented by the separation of PbI₂ into the ions Pb²⁺ and I⁻ when it is mixed with water. The reaction can be summarized by the equation:

• PbI₂ (s) ⇌ Pb²⁺ (aq) + 2 I⁻ (aq)

This reaction reaches saturation when no more solute can dissolve, meaning that solid Lead(II) Iodide is in equilibrium with the ions in solution. The dissolution process is pivotal in determining not only the solubility but also the practical applications of a substance. For this reaction, the solubility product constant (Ksp) quantifies the extent of PbI₂ dissolution.

During the dissolution of PbI₂, one mole of the solid dissolves to form one mole of Pb²⁺ ions and two moles of I⁻ ions. The Ksp expression,

• Ksp = [Pb²⁺][I⁻]²

is used to calculate how much of this compound can be dissolved in a given amount of solvent. This in turn influences various practical considerations, such as the effectiveness of reactions in solutions or the removal of lead from water. Understanding these aspects of dissolution is important for applications involving chemical reactions, waste treatment, and more.