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Chapter 15: Problem 44

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

Short Answer

Expert verified
The equilibrium concentrations are: \(Pb^{2+} = 1.58 \times 10^{-5} M\), \(Cr^{2+} = 3.16 \times 10^{-5} M\), and \(Cr^{3+} = 0.0999684 M\).

Step by step solution

01

Setting up the ICE table

Set up the ICE (Initial, Change, Equilibrium) table. Initially, the concentrations are \(Pb^{2+} =0M\), \(Cr^{2+} =0M\) and \(Cr^{3+} = 0.100M\). The change in concentrations are due to the chemical reaction. As a result, at equilibrium, the concentrations will be: \(Pb^{2+} = x\), \(Cr^{2+} = 2x\), and \(Cr^{3+} = 0.100 - 2x\). The reaction stoichiometry signifies that when one \(Pb\) atom reacts, it forms one \(Pb^{2+}\) ion and two \(Cr^{2+}\) ions, at the same time reducing two \(Cr^{3+}\) ions.
02

Equilibrium constant expression

Write the expression for the equilibrium constant. The equilibrium constant \(K_c\) for this reaction is defined as : \[K_c= \frac {[Pb^{2+}][Cr^{2+}]^{2}}{[Cr^{3+}]^{2}}\] Replace the equilibrium concentrations of the ions in the \(K_c\) expression. Thus, \(K_c= \frac {x(2x)^{2}}{(0.100-2x)^{2}}=3.2 \times 10^{-10}\].
03

Solving the equilibrium equation

Then, solve this quadratic equation for the unknown \(x\), which represents the equilibrium concentration of \(Pb^{2+}\). After solving the equation, \(x=1.58 \times 10^{-5}\). Thus the equilibrium concentrations are: \(Pb^{2+} = x=1.58 \times 10^{-5} M\), \(Cr^{2+} = 2x = 3.16 \times 10^{-5} M\), and \(Cr^{3+} = 0.100 - 2x = 0.0999684 M\).

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Most popular questions from this chapter

Use the following data to estimate a value of \(K_{\mathrm{p}}\) at \(1200 \mathrm{K}\) for the reaction \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) $$\begin{array}{l} \text { C(graphite) }+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \quad K_{\mathrm{c}}=0.64 \\ \quad \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \\ \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g}) \quad K_{\mathrm{c}}=1 \times 10^{8} \end{array}$$

In the gas phase, iodine reacts with cyclopentene \(\left(\mathrm{C}_{5} \mathrm{H}_{8}\right)\) by a free radical mechanism to produce cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) and hydrogen iodide. Explain how each of the following affects the amount of \(\mathrm{HI}(\mathrm{g})\) present in the equilibrium mixture in the reaction \begin{array}{r} \mathrm{I}_{2}(\mathrm{g})+\mathrm{C}_{5} \mathrm{H}_{8}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{HI}(\mathrm{g}) \\ \Delta H^{\circ}=92.5 \mathrm{kJ} \end{array} (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst; (e) adding an inert gas such as He to a constant-volume reaction mixture.

If the volume of an equilibrium mixture of \(\mathrm{N}_{2}(\mathrm{g}), \mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{NH}_{3}(\mathrm{g})\) is reduced by doubling the pressure, will \(P_{\mathrm{N}_{2}}\) have increased, decreased, or remained the same when equilibrium is re established? Explain. $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$$

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.
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