Problem 27
Question
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: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+}\) ?
Step-by-Step Solution
Verified Answer
The cation present in the unknown salt solution that forms a precipitate with all three mentioned solutions, $\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4}$, and $\mathrm{NaOH}$, is \(\boxed{\mathrm{Pb}^{2+}}\).
1Step 1: Write the general reactions with each solution and the solubility rules
For each cation, consider their precipitation reactions:
1. \(\mathrm{HBr}\) reaction: \[M_aX_b + a\mathrm{HBr} \rightarrow b\mathrm{HX} + aM\mathrm{Br}\]
2. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reaction: \[M_aX_b + ab\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow b\mathrm{HX} + aM(\mathrm{SO}_{4})\]
3. \(\mathrm{NaOH}\) reaction: \[M_aX_b + ab\mathrm{NaOH} \rightarrow b\mathrm{HX} + aM(\mathrm{OH})\]
Where \(M\) represents the unknown cation, \(\mathrm{HX}\) represents the unknown anion, and \(a\) and \(b\) are the stoichiometric coefficients.
Next, recall the solubility rules:
- Most alkali metal (Group 1) compounds are soluble.
- Most nitrate, acetate, and halide compounds are soluble, except for salts with silver, lead, or mercury(I) cations.
- Most sulfate compounds are soluble, with the main exceptions being barium, calcium, and lead sulfates.
- Most hydroxide salts are only slightly soluble, with the main exceptions being alkali metal hydroxides and heavy Group 2 metal hydroxides.
2Step 2: Analyze the solubility for each cation when combined with the three solutions
For each cation given, check the solubility with the mentioned compounds and determine if they form a precipitate in each case.
1. Potassium \(\mathrm{K}^{+}\):
- \(\mathrm{KBr}\) : soluble (alkali metal)
- \(\mathrm{K}_{2}\mathrm{SO}_{4}\) : soluble (alkali metal)
- \(\mathrm{KOH}\) : soluble (Group 1 and alkali metal)
2. Lead \(\mathrm{Pb}^{2+}\):
- \(\mathrm{PbBr}_{2}\) : insoluble (exceptions of solubility for halides)
- \(\mathrm{PbSO}_{4}\) : insoluble (exception of solubility for sulfates)
- \(\mathrm{Pb(OH)_{2}}\) : insoluble (only alkali metals are soluble exceptions for hydroxides)
3. Barium \(\mathrm{Ba}^{2+}\):
- \(\mathrm{BaBr}_{2}\) : soluble (no exceptions apply)
- \(\mathrm{BaSO}_{4}\) : insoluble (exception of solubility for sulfates)
- \(\mathrm{Ba(OH)}_{2}\) : insoluble (heavy Group 2 element, slightly soluble)
3Step 3: Determine which cation meets the criteria
Based on the solubility analysis above, the cation that forms a precipitate with all three mentioned solutions is \(\mathrm{Pb}^{2+}\), as it generates insoluble compounds with \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\).
Therefore, the cation present in the unknown salt solution is \(\boxed{\mathrm{Pb}^{2+}}\).
Key Concepts
Solubility RulesCation IdentificationLead (Pb²⁺) Precipitates
Solubility Rules
Understanding solubility rules is essential when predicting whether a precipitate will form during a chemical reaction. Solubility rules are guidelines that help determine the solubility of compounds in water. Here are some key solubility rules:
- Most alkali metals (such as sodium, potassium) and ammonium compounds are soluble.
- Nitrate (NO₃⁻), acetate (CH₃COO⁻), and most halide (Cl⁻, Br⁻, I⁻) salts are typically soluble, with notable exceptions being the halides of silver, lead, and mercury(I).
- Sulfates (SO₄²⁻) are generally soluble, except for those combined with barium, calcium, and lead.
- Hydroxides (OH⁻) are largely insoluble, with exceptions being alkali metals and certain heavier Group 2 metals like barium.
Cation Identification
Identifying cations based on observed precipitation reactions requires understanding how they interact with various anions. A cation is a positively charged ion that is fundamental in forming ionic compounds. When a solution contains a mixture of different ions, observing which ions form insoluble compounds can help identify them.
- The formation of a precipitate indicates that a new, less soluble compound has formed.
- This is often applied in analytical chemistry to detect the presence of specific ions in a solution.
- For example, treating a cation with \(\mathrm{HBr}\), \(\mathrm{H}_2\mathrm{SO}_4\), and \(\mathrm{NaOH}\) can indicate lead presence if precipitates form since these reactions align with the known insolubility of \(\mathrm{PbBr}_2\), \(\mathrm{PbSO}_4\), and \(\mathrm{Pb(OH)}_2\).
Lead (Pb²⁺) Precipitates
Lead (Pb²⁺) is a fascinating ion when studying precipitation reactions because it frequently forms insoluble compounds. In a solution analysis, identifying lead involves several key reactions:
- **Reaction with Bromides:** Lead bromide (\(\mathrm{PbBr}_2\)) is an example of a compound that forms a precipitate, demonstrating low solubility in water.
- **Reaction with Sulfates:** Lead sulfate (\(\mathrm{PbSO}_4\)) is another insoluble compound, aligning with sulfate solubility exceptions which include barium and lead.
- **Reaction with Hydroxides:** Lead hydroxide (\(\mathrm{Pb(OH)}_2\)) is also insoluble, complementing hydroxide solubility exceptions where lead is not soluble.
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