Chapter 18: Problem 39
When 1.55 g of solid thallium (I) bromide is added to \(1.00 \mathrm{L}\) of water, the salt dissolves to a small extent. $$\operatorname{TIBr}(\mathrm{s}) \rightleftarrows \mathrm{Tl}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})$$ The thallium(I) and bromide ions in equilibrium with TIBr each have a concentration of \(1.9 \times 10^{-3} \mathrm{M} .\) What is the value of \(K_{\mathrm{sp}}\) for TIBr?
Short Answer
Step by step solution
Understanding the Dissolution Process
Writing the Solubility Product Expression
Substituting Concentrations into \(K_{\text{sp}}\) Formula
Calculating the Value of \(K_{\text{sp}}\)
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thallium Bromide Dissolution
Equilibrium Concentrations
- At equilibrium, the concentration of \( \text{Tl}^+ \) is \(1.9 \times 10^{-3} \text{M}\).
- The concentration of \( \text{Br}^- \) is also \(1.9 \times 10^{-3} \text{M}\).
Ion Concentration Calculation
Chemical Equilibrium
- The system achieves balance, not because reactions stop, but because their rates equalize.
- The dissolving process results in a steady concentration of ions.