Problem 60 - Chemistry

Question

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. \(\left[K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_{2}(s) \text { is } 1.6 \times 10^{-5}.\right]\)

Step-by-Step Solution

Answer

The equilibrium concentrations of Pb²⁺ and Cl⁻ in the solution are approximately 0.050 M and 0.85 M, respectively.

Step 1: Calculate the initial moles of Pb²⁺ and Cl⁻.

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

Step 2: Use ICE table for concentration at equilibrium.

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.

Step 3: Apply the Ksp expression and find equilibrium concentrations of Pb²⁺ and Cl⁻.

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.

Key Concepts

Ksp (Solubility Product Constant)

The Solubility Product Constant, often represented as \( K_{sp} \), is a critical part of understanding solubility equilibria in chemistry. It reflects the product of the ion concentrations of a sparingly soluble compound at equilibrium in a saturated solution. In simpler terms, it's a way to describe how much of a solid can dissolve in water before it reaches its maximum solubility.Understanding \( K_{sp} \) is essential for predicting whether a solid will precipitate under certain conditions. It specifically applies to salts in an aqueous solution, like \( \text{PbCl}_2 \) in the given exercise. Each substance has a specific \( K_{sp} \) value, which varies depending on temperature and pressure. The smaller the \( K_{sp} \) value, the less soluble the compound. When calculating \( K_{sp} \) for a reaction such as the dissolution of \( \text{PbCl}_2 \), we express it as:- \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \)The equation shows that \( K_{sp} \) involves multiplying the concentrations of the ions at equilibrium, raised to the power of their stoichiometric coefficients in the balanced chemical equation. By comparing the ion product of a solution with its \( K_{sp} \), you can tell if the solution is unsaturated, saturated, or if precipitation will occur.

ICE Table

An ICE table is a tool used to track the initial conditions, changes, and equilibrium concentrations of species involved in a chemical reaction. The acronym "ICE" stands for Initial, Change, and Equilibrium. This table simplifies the process of solving equilibrium problems by organizing data in a structured way, which is particularly useful when working with reactions that involve changes in concentration.In the exercise presented, the ICE table helps to determine the concentrations of \( \text{Pb}^{2+} \) and \( \text{Cl}^- \) ions at equilibrium in the mixed solution. Here's how it breaks down:- **Initial Row:** Lists the initial moles of reactants. For example, \( \text{Pb}^{2+} \) is initially 0.005 mol and \( \text{Cl}^- \) is 0.050 mol.- **Change Row:** Quantifies how the molar concentrations change as the reaction reaches equilibrium. For instance, if \( x \) mol of \( \text{PbCl}_2 \) precipitates, \( -x \) mol of \( \text{Pb}^{2+} \) is removed, and \( -2x \) mol of \( \text{Cl}^- \) is removed because two chloride ions are required per lead ion.- **Equilibrium Row:** Gives the concentrations at equilibrium by adjusting the initial concentrations with the changes from the change row.Using the ICE table allows us to set up an equation with the \( K_{sp} \), facilitating the calculation of precise ion concentrations at equilibrium.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in an aqueous solution to form an insoluble product, or precipitate. These reactions are a key concept in understanding chemical equilibria and are particularly relevant in the provided exercise where \( \text{PbCl}_2 \) precipitates from the solution.In our given reaction scenario, combining \( \text{Pb(NO}_3)_2 \) and \( \text{KCl} \) in water leads to the following precipitation reaction:- \( \text{Pb}^{2+} \) from \( \text{Pb(NO}_3)_2 \), and \( \text{Cl}^- \) from \( \text{KCl} \) combine to form \( \text{PbCl}_2(s) \), a solid precipitate.Here's why and how it happens:- **Ion Collision:** When solutions containing these ions mix, ions collide randomly. If the product of their concentrations in solution exceeds their \( K_{sp} \), a solid precipitate forms until equilibrium is reached.- **Saturation Limit:** The \( K_{sp} \) acts as a limit of solubility. If the combined ion concentrations in the solution are greater than this limit, precipitation ensures the solution doesn't exceed saturation.The precipitate formation is not just a "reaction" step but a shift to equilibrium. In the exercise, after establishing ion concentrations via the ICE table, \( \text{PbCl}_2 \) reaches a point where no more of it can dissolve if the solution's ion concentration product exceeds \( K_{sp} \). This precipitate removal continues, ensuring equilibrium as defined by the \( K_{sp} \).