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

Short Answer

Expert verified
The equilibrium concentrations of Pb虏鈦 and Cl鈦 in the solution are approximately 0.050 M and 0.85 M, respectively.

Step by step solution

01

Calculate the initial moles of Pb虏鈦 and Cl鈦.

First, we need to find out the amount of moles for Pb虏鈦 and Cl鈦 ions from the volume and molar concentration provided. For Pb虏鈦: Moles of Pb虏鈦 = (Volume of Pb(NO鈧)鈧 脳 Molarity of Pb(NO鈧)鈧) Moles of Pb虏鈦 = (50.0 mL 脳 0.10 mol/L). Remember that we have to convert mL to L. Moles of Pb虏鈦 = (0.050 L 脳 0.10 mol/L) = 0.005 mol For Cl鈦: Moles of Cl鈦 = (Volume of KCl 脳 Molarity of KCl) Moles of Cl鈦 = (50.0 mL 脳 1.0 mol/L). Convert mL to L. Moles of Cl鈦 = (0.050 L 脳 1.0 mol/L) = 0.050 mol
02

Use ICE table for concentration at equilibrium.

Using an ICE (Initial, Change, Equilibrium) table, we will list the moles of Pb虏鈦 and Cl鈦 before mixing and their changes due to precipitation. Through this, we can track the equilibrium concentrations. | | Pb虏鈦 (aq) | + | 2 Cl鈦 (aq) | <=> | PbCl鈧 (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 Pb虏鈦 that reacts with Cl鈦 to form PbCl鈧.
03

Apply the Ksp expression and find equilibrium concentrations of Pb虏鈦 and Cl鈦.

The Ksp expression for this reaction is: Ksp = [Pb虏鈦篯 [Cl鈦籡虏 Given, Ksp = 1.6 脳 10鈦烩伒 At equilibrium, the concentrations of Pb虏鈦 and Cl鈦 are obtained by dividing their moles by the total volume (in liters). [Pb虏鈦篯 = \(\frac{0.005-x}{0.100}\) mol/L [Cl鈦籡 = \(\frac{0.050-2x}{0.100}\) mol/L Now, we can substitute the equilibrium concentrations in the Ksp expression: 1.6 脳 10鈦烩伒 = \(\frac{0.005-x}{0.100}\) 脳 \(\frac{0.050-2x}{0.100}\)虏 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: [Pb虏鈦篯 鈮 \(\frac{0.005}{0.100}\) mol/L 鈮 0.050 M [Cl鈦籡 鈮 \(\frac{0.050-2(0.005)}{0.100}\) mol/L 鈮 0.85 M Thus, the equilibrium concentrations of Pb虏鈦 and Cl鈦 in the solution are 0.050 M and 0.85 M, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( K_{sp} \).

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Most popular questions from this chapter

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M\) \(\mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)\)

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

If \(10.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) is added to 10.0 \(\mathrm{mL}\) of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-15}.\)

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0 \mathrm{MNH}_{3} ?\) $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ $$\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=5.9 \times 10^{-15}$$

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