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

First, we need to find out the amount of moles for Pb²⁺ and Cl⁻ ions from the volume and molar concentration provided. For Pb²⁺: Moles of Pb²⁺ = (Volume of Pb(NO₃)₂ × Molarity of Pb(NO₃)₂) Moles of Pb²⁺ = (50.0 mL × 0.10 mol/L). Remember that we have to convert mL to L. Moles of Pb²⁺ = (0.050 L × 0.10 mol/L) = 0.005 mol For Cl⁻: Moles of Cl⁻ = (Volume of KCl × Molarity of KCl) Moles of Cl⁻ = (50.0 mL × 1.0 mol/L). Convert mL to L. Moles of Cl⁻ = (0.050 L × 1.0 mol/L) = 0.050 mol

Using an ICE (Initial, Change, Equilibrium) table, we will list the moles of Pb²⁺ and Cl⁻ before mixing and their changes due to precipitation. Through this, we can track the equilibrium concentrations. | | Pb²⁺ (aq) | + | 2 Cl⁻ (aq) | <=> | PbCl₂ (s) | |---------------|-----------|---|-----------|-----|----------| | Initial (mol) | 0.005 | | 0.050 | | "x" | | Change (mol) | -"x" | | -2"x" | | +"x" | | Equilibrium | 0.005-"x" | | 0.050-2"x" | | "x" | Here, "x" represents the amount of Pb²⁺ that reacts with Cl⁻ to form PbCl₂.

The Ksp expression for this reaction is: Ksp = [Pb²⁺] [Cl⁻]² Given, Ksp = 1.6 × 10⁻⁵ At equilibrium, the concentrations of Pb²⁺ and Cl⁻ are obtained by dividing their moles by the total volume (in liters). [Pb²⁺] = \(\frac{0.005-x}{0.100}\) mol/L [Cl⁻] = \(\frac{0.050-2x}{0.100}\) mol/L Now, we can substitute the equilibrium concentrations in the Ksp expression: 1.6 × 10⁻⁵ = \(\frac{0.005-x}{0.100}\) × \(\frac{0.050-2x}{0.100}\)² Upon solving the equation for "x", we get that x is approximately equal to 0.005 (assuming x is much smaller than 0.005, and so the subtraction part of "0.005-x" can be ignored). So, at equilibrium, the concentrations are: [Pb²⁺] ≈ \(\frac{0.005}{0.100}\) mol/L ≈ 0.050 M [Cl⁻] ≈ \(\frac{0.050-2(0.005)}{0.100}\) mol/L ≈ 0.85 M Thus, the equilibrium concentrations of Pb²⁺ and Cl⁻ in the solution are 0.050 M and 0.85 M, respectively.