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: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+}\) ?

Short 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 \(\boxed{\mathrm{Pb}^{2+}}\).

Step-by-step solution

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.

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)

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 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.

These rules help deduce that - Compounds like - \(\mathrm{KBr}, \\mathrm{K}_2\mathrm{SO}_4\\mathrm{KOH}\) are soluble because potassium is an alkali metal. - Lead compounds \(\mathrm{PbBr}_2\), \(\mathrm{PbSO}_4\), \(\mathrm{Pb(OH)}_2\) are insoluble due to being exceptions. Understanding these exceptions is crucial when identifying cations like \(\mathrm{Pb}^{2+}\) that readily form precipitates.

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 \(\mathrm{Pb(OH)}_2\).

Through understanding cation behavior in these reactions, one can effectively identify elements like lead (Pb²⁺) within an unknown solution.

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.

These properties make lead an easy target for identification in an unknown salt solution. By observing reactions that lead to precipitation, one can conclude the presence of \(\mathrm{Pb}^{2+}\) efficiently. It's these distinct solubility characteristics that make lead a critical subject in chemical analysis.