Problem 44
Question
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? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$
Step-by-Step Solution
Verified Answer
The equilibrium concentrations are: \(Pb^{2+} = 1.58 \times 10^{-5} M\), \(Cr^{2+} = 3.16 \times 10^{-5} M\), and \(Cr^{3+} = 0.0999684 M\).
1Step 1: 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.
2Step 2: 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}\].
3Step 3: 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 \(Cr^{3+} = 0.100 - 2x = 0.0999684 M\).
Key Concepts
ICE TableEquilibrium ConstantStoichiometryReaction Stoichiometry
ICE Table
An ICE table is an essential tool in chemistry that helps track the concentrations of reactants and products over time.
It stands for Initial, Change, and Equilibrium.
Here's how each part works:
It stands for Initial, Change, and Equilibrium.
Here's how each part works:
- Initial: These are the concentrations at the start. For this problem, the initial concentrations are given as \(0M\) for \(Pb^{2+}\) and \(Cr^{2+}\), and \(0.100M\) for \(Cr^{3+}\).
- Change: This reflects how the concentrations change during the reaction. Here, since the reaction forms one \(Pb^{2+}\) and two \(Cr^{2+}\) for every two \(Cr^{3+}\) that react, the change is represented by \(+x, +2x\), and \(-2x\) respectively.
- Equilibrium: These are the concentrations when the reaction reaches equilibrium. They can be expressed as \(x\) for \(Pb^{2+}\), \(2x\) for \(Cr^{2+}\), and \(0.100-2x\) for \(Cr^{3+}\).
Equilibrium Constant
The equilibrium constant, \(K_c\), is a crucial value that reflects the balance between products and reactants at equilibrium.
For the reaction provided, the equilibrium constant expression is:\[K_c = \frac{[Pb^{2+}][Cr^{2+}]^{2}}{[Cr^{3+}]^{2}}\]This expression lets you understand how the concentrations at equilibrium relate to each other.
For the reaction provided, the equilibrium constant expression is:\[K_c = \frac{[Pb^{2+}][Cr^{2+}]^{2}}{[Cr^{3+}]^{2}}\]This expression lets you understand how the concentrations at equilibrium relate to each other.
- The key is that \(K_c\) only changes with temperature, not concentration changes or other physical conditions.
- It is calculated by plugging the equilibrium concentrations evaluated from the ICE table into the expression.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products.
It is derived from the balanced chemical equation.
It is derived from the balanced chemical equation.
- In this problem, the stoichiometric coefficients are 1 for \(Pb\), 2 for \(Cr^{3+}\), 1 for \(Pb^{2+}\), and 2 for \(Cr^{2+}\).
- This indicates that for every mole of \(Pb\) that reacts, 2 moles of \(Cr^{3+}\) are consumed, yielding 1 mole of \(Pb^{2+}\) and 2 moles of \(Cr^{2+}\).
Reaction Stoichiometry
Reaction stoichiometry refers to using stoichiometric coefficients to calculate concentration changes as shown in the ICE table.
This concept is especially useful in determining how changes in one species affect others.
This concept is especially useful in determining how changes in one species affect others.
- The ratio of coefficients gives you the relationships between the molecules involved in the reaction.
- For example, the stoichiometry between \(Cr^{3+}\) and \(Cr^{2+}\) is 2:2, indicating that as 2 \(Cr^{3+}\) ions are reduced, 2 \(Cr^{2+}\) ions form.
Other exercises in this chapter
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