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 \(25^{\circ} \mathrm{C}\).

Short answer

The solubility product constant (Ksp) for lead(II) iodide (PbI2) at 25°C is approximately \(6.43 \times 10^{-9}\).

Step-by-step solution

Calculate the molar concentration of PbI2

We are given that a 1.00 L saturated solution contains 0.54 g of PbI2. To find the molar concentration, we will first convert the mass of PbI2 to moles and then divide by the volume of the solution in liters. Molecular weight of PbI2 = 207.2 (Pb) + 2 * 126.9 (I) = 460.0 g/mol Number of moles of PbI2 = \(\frac{0.54 \mathrm{~g}}{460.0 \mathrm{~g/mol}}\) = \(1.17 \times 10^{-3} \mathrm{mol}\) Molar concentration of PbI2 = \(\frac{1.17 \times 10^{-3} \mathrm{mol}}{1.00 \mathrm{~L}}\) = \(1.17 \times 10^{-3} \mathrm{M}\)

Write the balanced chemical equation and solubility product expression

For lead(II) iodide, the balanced chemical equation is: \[ \mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\] The corresponding solubility product expression is: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2\]

Determine the concentrations of Pb²⁺ and I⁻ ions

Since the molar concentration of PbI2 is \(1.17 \times 10^{-3} \mathrm{M}\), the concentrations of the dissociated ions are as follows: - 1 mole of PbI2 gives 1 mole of Pb²⁺ ions: \([\mathrm{Pb}^{2+}] = 1.17 \times 10^{-3} \mathrm{M}\) - 1 mole of PbI2 gives 2 moles of I⁻ ions: \([\mathrm{I}^{-}] = 2 \times 1.17 \times 10^{-3} \mathrm{M} = 2.34 \times 10^{-3} \mathrm{M}\)

Calculate the solubility product constant (Ksp)

Now, we will use the solubility product expression with the calculated ion concentrations to find Ksp: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2 = (1.17 \times 10^{-3}) (2.34 \times 10^{-3})^2 = 6.43 \times 10^{-9} \] So, the solubility product constant for lead(II) iodide at 25°C is approximately \(6.43 \times 10^{-9}\).

Key concepts

Lead(II) Iodide Solubility

The solubility of lead(II) iodide (PbI₂) refers to the amount of this compound that can dissolve in a solvent, creating a saturated solution. In a saturated solution, the maximum amount of solute is dissolved at a given temperature and pressure.
Lead(II) iodide is slightly soluble in water, meaning only a small amount dissolves before equilibrium is reached between the dissolved ions and the undissolved solid.

• The dissolving process involves the dissociation of PbI₂ into its constituent ions, piece by piece, in the water.
• Because PbI₂ is composed of lead (Pb²⁺) and iodide (I⁻) ions, it dissolves by releasing these ions into the solution.

The solubility of a substance like lead(II) iodide can change with the temperature. Typically, increasing the temperature allows more PbI₂ to dissolve, as more energy is available to break the ionic bonds. Understanding solubility is crucial since it underpins many reactions and processes in chemistry.

Molar Concentration Calculation

Calculating molar concentration helps to quantify the amount of a solute (like PbI₂) in a given volume of solution. The solution's concentration indicates how much solute is present, which is key for predicting the behavior of solutes in reactions and solutions.
In the problem, we start by converting the mass of PbI₂ to moles since molarity is defined in terms of moles per liter.

• First, you calculate the molar mass of PbI₂ by adding the atomic masses of lead and iodide from the periodic table: 207.2 g/mol for lead and 126.9 g/mol for each iodide ion.


• The total molar mass is 460.0 g/mol for PbI₂.


• Next, divide the given mass by the molar mass: \( rac{0.54 ext{ g}}{460.0 ext{ g/mol}} = 1.17 imes 10^{-3} ext{ mol} \).


• Finally, to find the molar concentration or molarity, divide the moles by the volume of the solution (in liters): \( rac{1.17 imes 10^{-3} ext{ mol}}{1.00 ext{ L}} = 1.17 imes 10^{-3} ext{ M} \).

Knowing the molar concentration is instrumental for determining the solubility product, helping to predict the solubility and behavior of ions in equilibrium.

Solubility Equilibrium

Solubility equilibrium is a dynamic state established when the rate of PbI₂ dissolving equals the rate of Pb²⁺ and I⁻ ions recombining into the solid. This balance allows us to define the solubility product constant, or Ksp.
In the case of lead(II) iodide, the equilibrium can be represented by the equation: \(\mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\).

• Ksp is a way to express the concentrations of the ions in a saturated solution at equilibrium directly through their concentrations.


• For PbI₂, the expression is \( K_{sp} = [Pb^{2+}] [I^{-}]^2 \), showing how each ion contributes to the equilibrium constant.


• The equilibrium expression indicates that a change in the concentration of either ion will impact the system until a new equilibrium is reached, maintaining the value of Ksp consistent under constant conditions.

By calculating Ksp, we can derive meaningful insights into the extent to which PbI₂ can dissolve, a fundamental part of solubility and precipitation reactions.