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

Expert verified
The solubility-product constant (Ksp) for lead(II) iodide (PbI₂) at 25°C is calculated by first finding the molar concentration of Pb²⁺ and I⁻ ions in the saturated solution. In this case, we have [Pb²⁺] = 0.00117 M and [I⁻] = 0.00234 M. Using the Ksp expression, Ksp = [Pb²⁺][I⁻]², we find that Ksp = \(6.43 * 10^{-9} M^3\).

Step by step solution

01

Determine the number of moles of PbI₂ in 1.00 L of solution

First, we will convert the given mass of PbI₂ (0.54 g) into moles by using the molar mass of PbI₂. To find the molar mass, add the atomic mass of lead and twice the atomic mass of iodine: PbI₂ molar mass = 207.2 g/mol (lead) + 2 * 126.9 g/mol (iodine) = 460.98 g/mol Now, divide the mass of PbI₂ in the solution by its molar mass: Moles of PbI₂ = (0.54 g) / (460.98 g/mol) = 0.00117 mol Since we have 1.00 L of solution, the molar concentration of PbI₂ is: [PbI₂] = (0.00117 mol) / (1.00 L) = 0.00117 M
02

Write the chemical equation for the dissolution of PbI₂

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)
03

Determine the molar concentration of Pb²⁺ and I⁻ ions

Since every mole of PbI₂ that dissolves produces one mole of Pb²⁺ ions and two moles of I⁻ ions, we have: [Pb²⁺] = 0.00117 M (from the dissociation of PbI₂) [2I⁻] = 2 * 0.00117 M = 0.00234 M
04

Calculate the solubility-product constant (Ksp) for PbI₂

We can now calculate the Ksp for PbI₂ using the Ksp expression: Ksp = [Pb²⁺] [I⁻]² Substitute the molar concentrations of Pb²⁺ and I⁻ ions: Ksp = (0.00117 M) * (0.00234 M)² = 6.43 * 10⁻⁹ M³ The solubility-product constant for lead(II) iodide at 25°C is 6.43 * 10⁻⁹ M³.

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Most popular questions from this chapter

Which of the following solutions is a buffer? (a) \(0.20 \mathrm{M}\) for\(\operatorname{mic}\) acid \((\mathrm{HCOOH}),(\mathbf{b}) 0.20 M\) formic acid \((\mathrm{HCOOH})\) and \(0.20 \mathrm{M}\) sodium formate \((\mathrm{HCOONa}),(\mathbf{c}) 0.20 \mathrm{Mnitric}\) acid \(\left(\mathrm{HNO}_{3}\right)\) and \(0.20 \mathrm{M}\) sodium nitrate \(\left(\mathrm{NaNO}_{3}\right)\) (d) both b and \(\mathrm{c},(\mathbf{e})\) all of \(\mathrm{a}, \mathrm{b},\) and \(\mathrm{c}\).

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For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

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