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 (Kₚ) for lead(II) iodide at 25°C is 6.40 x 10⁻⁹.

Step-by-step solution

Calculate the molar solubility of PbI₂

To find the molar solubility, we first need to calculate the number of moles of PbI₂ present in the 1.00-L solution. Given mass of PbI₂ = 0.54 g Molar mass of PbI₂ = (Pb=207.2 g/mol) + 2*(I=126.9 g/mol) = 207.2 + (2*126.9) = 460.99 g/mol Moles of PbI₂ = \(\frac{0.54 g}{460.99 g/mol}\) = \(1.17 \times 10^{-3} mol\) Now, we have the number of moles of PbI₂. We can find the molar solubility of PbI₂ in the solution by dividing the moles of PbI₂ by the volume of the solution. Molar solubility of PbI₂ = \(\frac{1.17 \times 10^{-3} mol}{1.00 L}\) = 1.17 x 10⁻³ mol/L

Determine the Balance Dissociation Equation for PbI₂

The balanced dissociation equation for PbI₂ can be written as: PbI₂(s) \(\rightleftharpoons\) Pb²⁺(aq) + 2I⁻(aq) This balanced equation tells us that for every one mole of Pb²⁺ in the solution, there will also be two moles of I⁻.

Calculating Concentration of Ions

We know that the molar solubility of PbI₂ is 1.17 x 10⁻³ mol/L, which is equal to the concentration of Pb²⁺ in the solution. Concentration of Pb²⁺ = 1.17 x 10⁻³ mol/L Since for every mole of Pb²⁺ there are two moles of I⁻, the concentration of I⁻ will be: Concentration of I⁻ = 2 x (1.17 x 10⁻³) = 2.34 x 10⁻³ mol/L

Calculate the solubility product constant (Kₚ)

Now that we have the concentration of Pb²⁺ and I⁻, we can find the solubility product constant Kₚ using the following expression: Kₚ = [Pb²⁺] × [I⁻]² Plugging in the values, we get: Kₚ = (1.17 x 10⁻³) × (2.34 x 10⁻³)² = 6.40 x 10⁻⁹ Thus, the solubility product constant for lead(II) iodide at 25°C is 6.40 x 10⁻⁹.