Problem 27 - Chemistry The Central Science In Si Units

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

Answer

The cation that could be present is Pb2+.

Step 1: Evaluate Precipitation with HBr

First, we need to consider the precipitation reaction that occurs when HBr reacts with the cations. We know that lead bromide (PbBr2) is an insoluble salt, which means it will precipitate when Pb2+ ions are present in the solution with bromide ions from HBr. Potassium bromide (KBr) and barium bromide (BaBr2) are soluble in water, so they would not form precipitates. Hence, a precipitate forming with HBr indicates the possible presence of Pb2+.

Step 2: Evaluate Precipitation with H2SO4

Next, we test for precipitates with H2SO4. Barium sulfate (BaSO4) is highly insoluble, so a precipitate will form if Ba2+ ions are present. Potassium sulfate (K2SO4) and lead sulfate (PbSO4) are relatively soluble in water, so they are less likely to precipitate under normal conditions. Therefore, a precipitate forming with H2SO4 indicates Ba2+ might be present.

Step 3: Evaluate Precipitation with NaOH

Finally, consider the reaction with NaOH. Lead hydroxide (Pb(OH)2) is insoluble in water, whereas barium hydroxide (Ba(OH)2) is more soluble. Potassium hydroxide is also soluble in water. Therefore, a precipitate forming with NaOH suggests the presence of Pb2+ ions.

Step 4: Conclusion on the Cations Present

Comparing results from the three reactions, Pb2+ forms a precipitate with HBr and NaOH and possibly even with H2SO4 (due to low solubility at high concentrations). Ba2+ only forms a precipitate with H2SO4. Since a precipitate formed in all cases, Pb2+ is the common ion that matches all outcomes.

Key Concepts

Precipitation Reactions

Precipitation reactions are a fascinating type of chemical reaction. They occur when two aqueous solutions are mixed and an insoluble solid, known as a precipitate, forms. This solid separates from the mixture and often appears as a cloudy or solid substance within the liquid. The key to these reactions is the interaction between the ions present in the solutions. When certain cations and anions come together, they may form a compound that does not dissolve in water.
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

Solubility rules provide guidance on whether a salt will dissolve in water. These rules are handy, especially when predicting if a precipitate will form during a reaction. Generally, most nitrate (\( \text{NO}_3^- \)) salts and alkali metal (\( \text{K}^+, \text{Na}^+ \)) salts are soluble. However, there are exceptions, mostly involving heavy metal ions and specific anions.Bullets of insoluble combinations:

• 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 (\( \text{KOH} \)).

Understanding these rules helps predict the outcomes of mixing various chemical solutions. In our exercise, these rules help determine which cations form precipitates when treated with \( \text{HBr} \), \( \text{H}_2\text{SO}_4 \), and \( \text{NaOH} \).

Unknown Salt Identification

Identifying an unknown salt involves applying the concepts of precipitation reactions and solubility rules. This analytical technique is used by chemists to determine the composition of a substance simply by observing which precipitates form upon adding specific reagents.The original problem involves treating an unknown salt solution with different reagents to see which cations are present. For instance:

• 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.

By systematically using these strategies alongside solubility rules, chemists can identify the components of an unknown sample effectively. This method is fundamental in analytical chemistry, teaching students essential techniques for solving real-world problems.