Question

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-} ; \mathrm{CO}_{3}^{2-} ; \mathrm{NO}_{3}^{-} ?\)

Short answer

The anion of the unknown salt is \(\mathrm{CO}_{3}^{2-}\) because it forms insoluble compounds with \(\mathrm{Ag}^{+}, \mathrm{Pb}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, consistent with the observation of precipitates in all cases.

Step-by-step solution

When combined with \(\mathrm{Ag}^{+}, \mathrm{Pb}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, the bromide ion \(\mathrm{Br}^{-}\) forms: 1. \(\mathrm{AgBr}\): Silver bromide, which is insoluble in water. 2. \(\mathrm{PbBr}_{2}\): Lead bromide, which is soluble in water. 3. \(\mathrm{BaBr}_{2}\): Barium bromide, which is soluble in water. Since precipitates are formed in all three cases, and only one of the results with bromide ions forms an insoluble compound, this cannot be the anion of the unknown salt. #Step 2: Solubility for \(\mathrm{CO}_{3}^{2-}\) compounds#

When combined with \(\mathrm{Ag}^{+}, \mathrm{Pb}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, the carbonate ion \(\mathrm{CO}_{3}^{2-}\) forms: 1. \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\): Silver carbonate, which is insoluble in water. 2. \(\mathrm{PbCO}_{3}\): Lead carbonate, which is insoluble in water. 3. \(\mathrm{BaCO}_{3}\): Barium carbonate, which is insoluble in water. Since precipitates form in all three cases and all the compounds formed by the carbonate ion are insoluble in water, this anion is a strong candidate for the anion of the unknown salt. #Step 3: Solubility for \(\mathrm{NO}_{3}^{-}\) compounds#

When combined with \(\mathrm{Ag}^{+}, \mathrm{Pb}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, the nitrate ion \(\mathrm{NO}_{3}^{-}\) forms: 1. \(\mathrm{AgNO}_{3}\): Silver nitrate, which is soluble in water. 2. \(\mathrm{Pb(NO}_{3})_{2}\): Lead nitrate, which is soluble in water. 3. \(\mathrm{Ba(NO}_{3})_{2}\): Barium nitrate, which is soluble in water. Since precipitates are formed in all three cases and none of the compounds formed by the nitrate ion are insoluble in water, this cannot be the anion of the unknown salt. #Conclusion#

Out of the given options \(\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\), the anion that forms insoluble compounds with \(\mathrm{Ag}^{+}, \mathrm{Pb}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions is \(\mathrm{CO}_{3}^{2-}\). Therefore, the anion of the unknown salt is \(\mathrm{CO}_{3}^{2-}\).

Key concepts

Solubility Rules

Understanding solubility rules is crucial when dealing with ionic compounds. These rules help predict whether a compound will dissolve in water or form a precipitate. For ionic compounds, the solubility depends on the combination of cations (positive ions) and anions (negative ions).

Certain ions have specific solubility characteristics. For example, nitrate ions (\( \text{NO}_3^- \)) are almost always soluble, meaning compounds containing these ions are highly likely to dissolve in water. On the other hand, for anions like carbonate (\( \text{CO}_3^{2-} \)), solubility depends heavily on the cation they are paired with.

Briefly, the solubility rules can be summarized as follows:

• Most nitrate (\( \text{NO}_3^- \)) and acetate (\( \text{CH}_3\text{COO}^- \)) salts are soluble.
• Salts containing Group 1 elements (Li+, Na+, K+, Cs+, Rb+) are generally soluble.
• Most carbonate (\( \text{CO}_3^{2-} \)) and phosphate (\( \text{PO}_4^{3-} \)) salts are insoluble, except when combined with alkali metals or ammonium ions.
• Halide salts (chlorides, bromides, iodides) are generally soluble, except for those of silver, lead, and mercury.

Knowing these rules allows us to determine whether a reaction will occur when ions are mixed in solution.

Precipitation Reactions

Precipitation reactions occur when two solutions containing soluble salts are mixed together, resulting in the formation of an insoluble compound. The insoluble compound, known as a precipitate, forms because the combination of ions in solution creates a new ionic compound that does not dissolve in water.

These reactions are represented by balanced chemical equations. For instance, when aqueous solutions of silver nitrate (\( \text{AgNO}_3 \)) and sodium chloride (\( \text{NaCl} \)) are combined, silver chloride (\( \text{AgCl} \)), an insoluble compound, forms as a white precipitate: \[ \text{Ag}^{+} (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) \]

It's important to note the conditions under which precipitation occurs:

• The newly formed ionic compound must be less soluble in the solvent than the reactants.
• The ions from different reactants exchange partners, usually resulting in the formation of at least one insoluble product.

Understanding precipitation reactions helps in predicting the products of a given chemical reaction and is fundamental in identifying ionic compounds in analytical chemistry.

Anion Identification

Anion identification is a process used to determine the negative ion in an unknown ionic compound. This determination can be made using chemical tests that rely on the solubility and precipitation characteristics of anions.

In the exercise presented, reactions with common ions such as \( \text{Ag}^+ \), (\( \text{Pb}^{2+} \)), and (\( \text{Ba}^{2+} \)) are used to identify the possible anion.

Here's how you can summarize the steps for identification:

• Mixing a solution of the unknown compound with one and then another of the test reagents: \( \text{AgNO}_3 \), (\( \text{Pb(NO}_3)_{2} \), and (\( \text{BaCl}_2 \)).
• Observing whether a precipitate forms in each scenario.

In the case given, (\( \text{CO}_3^{2-} \)) is identified as the anion because it forms insoluble compounds with all tested cations, while (\( \text{Br}^- \)) and (\( \text{NO}_3^{-} \)) do not. This method is a systematic approach to determine which anion could be present based on known chemical behavior.