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) of lead(II) iodide (\(PbI_2\)) at \(25^{\circ} \mathrm{C}\) can be calculated by first finding its molar solubility (0.00117 mol/L), then using the balanced chemical equation to determine the concentrations of the ions in the solution ([Pb²⁺] = 0.00117 mol/L, [I⁻] = 0.00234 mol/L), and finally applying the formula Ksp = [Pb²⁺][I⁻]². The Ksp for \(PbI_2\) at \(25^{\circ} \mathrm{C}\) is 6.44 x 10⁻⁶.

Step-by-step solution

Calculate the molar solubility of PbI₂

To find the molar solubility of PbI₂, we first need to convert the 0.54g of PbI₂ to moles. We can do this using its molar mass, which can be found by adding 207.2 g/mol (for lead) and 253.8 g/mol (for two iodine atoms). Molar mass of PbI₂ = 207.2 g/mol + 253.8 g/mol = 461.0 g/mol Now, we can find the number of moles of PbI₂: Moles of PbI₂ = (0.54 g) / (461.0 g/mol) = 0.00117 mol Since the volume of the solution is 1.00 L, the molar solubility (S) will be equal to the number of moles of PbI₂: S = 0.00117 mol/L

Write the balanced chemical equation for the dissolution of PbI₂

The balanced chemical equation for the dissolution of PbI₂ is: \(PbI_{2} \rightleftharpoons Pb^{2+} + 2I^-\) When 1 mole of lead(II) iodide dissolves in water, it produces 1 mole of lead(II) ions (Pb²⁺) and 2 moles of iodide ions (I⁻).

Calculate the concentrations of Pb²⁺ and I⁻ ions

From the balanced chemical equation, we can see that: 1 mole of PbI₂ → 1 mole of Pb²⁺ and 2 moles of I⁻ 0.00117 mol/L of PbI₂ → 0.00117 mol/L of Pb²⁺ and 0.00234 mol/L of I⁻ Now we have the concentrations of both ions in the solution: [Pb²⁺] = 0.00117 mol/L [I⁻] = 0.00234 mol/L

Calculate the solubility product constant (Ksp)

Now that we have the concentrations of Pb²⁺ and I⁻ ions, we can find the solubility product constant (Ksp) using the formula: Ksp = [Pb²⁺][I⁻]² Ksp = (0.00117)(0.00234)² = 6.44 x 10⁻⁶ So, the solubility product constant for lead(II) iodide (PbI₂) at 25°C is 6.44 x 10⁻⁶.