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

Short Answer

Expert verified
The equilibrium concentrations of Pb²⁺ and Cl⁻ ions are approximately 0.09984 M and 0.99968 M, respectively.

Step by step solution

01

Write the balanced chemical equation.

The balanced chemical equation for the dissolution of PbCl₂ in water is as follows: \(PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)\)
02

Set up the ICE table.

Here, we set up the ICE table to represent the initial concentrations of reactants and products, the change in concentrations (due to the reaction), and the equilibrium concentrations of ions: Initial concentrations: - [Pb²⁺] = 0.10 M (from Pb(NO₃)₂ solution) - [Cl⁻] = 1.0 M (from KCl solution) Change in concentrations: - [Pb²⁺] = -x - [Cl⁻] = -2x Equilibrium concentrations: - [Pb²⁺] = 0.10 - x - [Cl⁻] = 1.0 - 2x
03

Write the expression for the Ksp and solve for x.

The expression for the Ksp of PbCl₂ is: \(K_{sp} = [Pb^{2+}][Cl^{-}]^{2}\) Plugging in the equilibrium concentrations from the ICE table: \(K_{sp} = (0.10 - x)(1.0 - 2x)^{2}\) \(1.6 \times 10^{-5} = (0.10 - x)(1.0 - 2x)^{2}\) To solve for x, we can first make a simplifying assumption to make the calculations easier. We can assume that x is very small relative to 0.10 and 1.0, so we approximate (0.10 - x) ≈ 0.10 and (1.0 - 2x) ≈ 1.0. Now, we can solve for x: \(1.6 \times 10^{-5} = (0.10)(1.0)^{2}\) \(x = 1.6 \times 10^{-4}\) Since x is the change in concentrations of Pb²⁺ and Cl⁻ ions, we can now calculate the equilibrium concentrations: [Pb²⁺] = 0.10 - x = 0.10 - 1.6 x 10⁻⁴ ≈ 0.09984 M [Cl⁻] = 1.0 - 2x = 1.0 - 2(1.6 x 10⁻⁴) ≈ 0.99968 M At equilibrium, the concentrations of Pb²⁺ and Cl⁻ ions are approximately 0.09984 M and 0.99968 M, respectively.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial when delving into the behavior of reactions in a closed system. Equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of the reactants and products remain constant over time, but they are not necessarily equal.

For example, when lead(II) nitrate and potassium chloride are mixed, it results in the formation of lead(II) chloride. In this reaction, lead(II) chloride can both dissolve to produce ions and precipitate out of the solution. At equilibrium, these two processes occur at the same rate, meaning that the amount of lead(II) chloride in the solid and dissolved state does not change.

A key aspect of chemical equilibrium is that it's a dynamic process. Even though the macroscopic properties stay the same (like concentrations), at the microscopic level, individual molecules are continuously reacting in both the forward and reverse directions.
Ksp (Solubility Product Constant)
The Ksp, or solubility product constant, is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. It's a crucial concept for understanding the solubility and precipitation processes.

For the reaction where a salt, like PbCl₂, dissolves in water, the Ksp expression takes the form of the product of the concentrations of the resulting ions, each raised to the power of its stoichiometric coefficient. In our given exercise, the Ksp expression for PbCl₂ is represented as:
\[K_{sp} = [Pb^{2+}][Cl^{-}]^{2}\]

The Ksp value is constant at a given temperature and reflects the extent to which a compound can dissolve in water. If the product of the ionic concentrations in a solution exceeds the Ksp value, the excess ions will start to precipitate until the product of their concentrations equals the Ksp again. Conversely, if the product is below the Ksp value, more salt can dissolve until equilibrium is reestablished.
ICE Table Method
ICE stands for Initial, Change, and Equilibrium, and the method involves setting up a table to systematically calculate the equilibrium concentrations of all species in a reaction. It's a visual tool that helps clarify the step-by-step changes occurring in a reaction mixture.

Using an ICE table starts with noting the initial concentrations of reactants and products prior to any reaction occurring. Then, you denote the changes in concentrations as the reaction proceeds towards equilibrium, typically represented by variables like 'x'. Finally, you calculate the equilibrium concentrations.
For our problem where lead chloride dissolves in water:
  • Initial concentrations of Pb²⁺ and Cl⁻ are known from the Pb(NO₃)₂ and KCl solutions provided.
  • The change in concentrations is expressed in terms of 'x', corresponding to the number of moles of PbCl₂ that dissolve.
  • Equilibrium concentrations involve the initial concentrations adjusted by these changes (subtracting 'x' for Pb²⁺ and '2x' for Cl⁻ due to the stoichiometry).
The ICE table provides a simple yet powerful way to visualize and solve for equilibrium conditions, especially when dealing with solubility equilibria as illustrated in our exercise.

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Most popular questions from this chapter

Consider \(1.0 \mathrm{~L}\) of an aqueous solution that contains \(0.10 \mathrm{M}\) sulfuric acid to which \(0.30\) mol barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH}\), the concentration of barium ions in the final solution, and the mass of solid formed.

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

a. Calculate the molar solubility of AgBr in pure water. \(K_{\text {sp }}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) b. Calculate the molar solubility of AgBr in \(3.0 M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\), that is, \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}\) c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{~mL}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3}\) ? e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} M .\) Calculate \(K_{\text {sp }}\) for \(\mathrm{PbBr}_{2}\).

Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?

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