Chapter 4: Problem 27
Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+}\) ?
Short Answer
Step by step solution
Evaluate Precipitation with HBr
Evaluate Precipitation with H2SO4
Evaluate Precipitation with NaOH
Conclusion on the Cations Present
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Precipitation Reactions
For example, in our original exercise, when lead ions (\( \text{Pb}^{2+} \)) are introduced to bromide ions (\( \text{Br}^- \)) from hydrogen bromide (\( \text{HBr} \)), they form lead bromide (\( \text{PbBr}_2 \)), a highly insoluble salt. This results in visible precipitation.Understanding precipitation reactions can help in the identification of ions in a solution and can be used in various practical applications such as water treatment or in processes to remove unwanted ions from mixtures.
Solubility Rules
- Sulfates: Most are soluble, but lead sulfate (\( \text{PbSO}_4 \)) and barium sulfate (\( \text{BaSO}_4 \)) are exceptions and generally form precipitates.
- Halides (like chlorides, bromides): Mostly soluble, but those of lead (\( \text{PbBr}_2 \)) often form precipitates.
- Hydroxides: Usually insoluble, except for those of some Group 1 elements like potassium (\( \text{KOH} \)).
Unknown Salt Identification
- When a precipitate forms with \( \text{HBr} \), it suggests that lead ions (\( \text{Pb}^{2+} \)) may be present due to the formation of insoluble lead bromide.
- If a precipitate forms with \( \text{H}_2\text{SO}_4 \), barium ions (\( \text{Ba}^{2+} \)) might be involved due to barium sulfate's low solubility.
- A reaction with \( \text{NaOH} \) leading to a precipitate again points toward lead ions creating lead hydroxide, which is not soluble.