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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 Pb虏鈦 and Cl鈦 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 PbCl鈧 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: - [Pb虏鈦篯 = 0.10 M (from Pb(NO鈧)鈧 solution) - [Cl鈦籡 = 1.0 M (from KCl solution) Change in concentrations: - [Pb虏鈦篯 = -x - [Cl鈦籡 = -2x Equilibrium concentrations: - [Pb虏鈦篯 = 0.10 - x - [Cl鈦籡 = 1.0 - 2x
03

Write the expression for the Ksp and solve for x.

The expression for the Ksp of PbCl鈧 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) 鈮 0.10 and (1.0 - 2x) 鈮 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 Pb虏鈦 and Cl鈦 ions, we can now calculate the equilibrium concentrations: [Pb虏鈦篯 = 0.10 - x = 0.10 - 1.6 x 10鈦烩伌 鈮 0.09984 M [Cl鈦籡 = 1.0 - 2x = 1.0 - 2(1.6 x 10鈦烩伌) 鈮 0.99968 M At equilibrium, the concentrations of Pb虏鈦 and Cl鈦 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 PbCl鈧, 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 PbCl鈧 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 Pb虏鈦 and Cl鈦 are known from the Pb(NO鈧)鈧 and KCl solutions provided.
  • The change in concentrations is expressed in terms of 'x', corresponding to the number of moles of PbCl鈧 that dissolve.
  • Equilibrium concentrations involve the initial concentrations adjusted by these changes (subtracting 'x' for Pb虏鈦 and '2x' for Cl鈦 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

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}{ }^{2-}\).)

Calculate the mass of manganese hydroxide present in \(1300 \mathrm{~mL}\) of a saturated manganese hydroxide solution. For \(\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=\) \(2.0 \times 10^{-13}\) 89\. On a hot day, a \(200.0-\mathrm{mL}\) sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If \(240 \mathrm{mg}\) of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

A solution is prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-2} M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{NaF}\). Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-8}\right)\) precipitate?

The \(K_{\text {sp }}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). At what \(\mathrm{pH}\) will a \(0.2 \mathrm{M} \mathrm{Al}^{3+}\) solution begin to show precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) ?

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\left(K_{\mathrm{sp}}=4 \times\right.\) \(10^{-33}\) ) in each of the following. a. water b. a solution buffered at \(\mathrm{pH}=5.0\) c. a solution buffered at \(\mathrm{pH}=11.0\)

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