Chapter 23: Problem 67
Find the concentrations of \(\mathrm{Cu}^{2+}(a q), \mathrm{NH}_{3}(a q)\), and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) at equilibrium when \(0.10 \mathrm{~mol} \mathrm{Cu}^{2+}(a q)\) and \(0.40 \mathrm{~mol} \mathrm{NH}_{3}(a q)\) are made up to \(1.00 \mathrm{~L}\) of solution. The dissociation constant, \(K_{d}\), for the complex \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is \(2.1 \times 10^{-13}\)
Short Answer
Step by step solution
Write the Equilibrium Reaction
Define Initial Concentrations
Express Changes Using x at Equilibrium
Use the Dissociation Constant
Make an Assumption to Simplify Calculations
Solve for x
Calculate Equilibrium Concentrations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Concentrations
- \([\mathrm{Cu}^{2+}] = 0.10 - x\)
- \([\mathrm{NH}_3] = 0.40 - 4x\)
- \([\mathrm{Cu(NH}_3)_4^{2+}] = x\)
Dissociation Constant
Complex Ions
- Their ability to distribute charge over a larger volume, affecting stability and reactivity.
- Formation of specific geometric configurations around the central metal atom.
- Significant role in a variety of chemical reactions, especially in catalysis and material science.
Equilibrium Calculations
- \([\text{Cu}^{2+}] \approx 0.10 \text{ M}\)
- \([\text{NH}_3] \approx 0.40 \text{ M}\)
- \([\text{Cu(NH}_3\text{)}_4^{2+}] \approx 1.22 \times 10^{-14} \text{ M}\)