A solution is prepared by mixing \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{~mL}\) of \(1.0 M \mathrm{KCl}\). Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\mathrm{sp}}\right.\) for \(\mathrm{PbCl}_{2}(s)\) is \(\left.1.6 \times 10^{-5} .\right]\)

First, we need to write the balanced chemical equation for the reaction when the two solutions are mixed. The reaction involves the formation of solid lead(II) chloride (PbCl₂) from aqueous lead(II) nitrate (Pb(NO₃)₂) and aqueous potassium chloride (KCl): Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

Next, we need to find the initial moles of both reactants in the mixture. We will first determine the moles of Pb²⁺ ions and Cl⁻ ions. For Pb²⁺ ions: Moles = Volume × Molarity Moles = \(50.0\:mL × 0.10\:M\) Moles = \(5.0 \times 10^{-3}\:mol\) For Cl⁻ ions: Moles = Volume × Molarity Moles = \(50.0\:mL × 1.0\:M\) Moles = \(50.0 \times 10^{-3}\:mol\)

Now, we need to determine how much Pb²⁺ and Cl⁻ ions reacted according to stoichiometry. According to the balanced equation, 1 mole of Pb²⁺ ions reacts with 2 moles of Cl⁻ ions. Therefore, both moles will not be consumed completely. From the stoichiometry of the reaction, we can calculate the moles of Cl⁻ ions that reacted with Pb²⁺ ions: Moles of Cl⁻ ions used = 2 × moles of Pb²⁺ ions Moles of Cl⁻ ions used = 2 × \(5.0 \times 10^{-3}\:mol\) Moles of Cl⁻ ions used = \(10.0 \times 10^{-3}\:mol\) After the reaction occurs, we will have the following moles: Moles of Cl⁻ ions remaining = initial moles of Cl⁻ ions - moles of Cl⁻ ions used Moles of Cl⁻ ions remaining = \((50.0 - 10.0) \times 10^{-3}\:mol\) Moles of Cl⁻ ions remaining = \(40.0 \times 10^{-3}\:mol\) Now, we have no moles of Pb²⁺ ions remaining in solution since it has all precipitated as PbCl2. Next, we need to find the equilibrium concentrations using the total volume of the solution, which is 100 mL: \[Equilibrium\_Concentration\ of [Cl^-] = \frac{Moles\ of\ Cl^- \:ions}{Volume\ of\ solution}\] \[Equilibrium\_Concentration\ of [Cl^-] = \frac{40.0 \times 10^{-3}\:mol}{100.0\:mL}\] \[Equilibrium\_Concentration\ of [Cl^-] = 0.4\:M\] The concentration of Pb²⁺ ions in the solution can be found using Ksp: \[K_{sp} = [Pb^{2+}] [Cl^-]^2\] \[1.6 \times 10^{-5} = [Pb^{2+}] (0.4)^2\] Now, solve for [Pb²⁺]: \[ [Pb^{2+}] = \frac{1.6 \times 10^{-5}}{(0.4)^2}\] \[ [Pb^{2+}] = 1.0 \times 10^{-5}\]

Now, we have the equilibrium concentrations of Pb²⁺ and Cl⁻ ions in the solution: [Cl⁻] = 0.4 M [Pb²⁺] = \(1.0 \times 10^{-5}\:M\)