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Chapter 17: Problem 54

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

Expert verified
The solubility-product constant (Ksp) for PbI2 at 25°C is \( 6.41 \times 10^{-9} \).

Step by step solution

01

Write the balanced dissociation equation for PbI2

The balanced dissociation equation for PbI2 in water is: \[ \mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)} \]
02

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: \[ \mathrm{K_{sp} = [Pb^{2+}][I^-]^2} \]
03

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: \[ [\mathrm{Pb^{2+}}] = 0.00117 \,\mathrm{M} \] \[ [\mathrm{I^-}] = 2 \times 0.00117 \,\mathrm{M} = 0.00234 \,\mathrm{M} \]
04

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 \( 6.41 \times 10^{-9} \).

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