Problem 66 - Chemistry An Atoms First Approach

Question

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI} ?\)

Step-by-Step Solution

Answer

In this exercise, we analyzed a series of chemical reactions involving silver halides. The reactions and their balanced chemical equations are as follows: 1. \(AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3\) 2. \(AgCl_{(s)} + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\) 3. \([Ag(NH_3)_2]^+ + Cl^- + KBr \rightarrow AgBr_{(s)} + 2NH_3 + KCl\) 4. \(AgBr_{(s)} + 2Na_2S_2O_3 \rightarrow [Ag(S_2O_3)_2]^{3-} + 2NaBr\) 5. \([Ag(S_2O_3)_2]^{3-} + 2NaBr + 2KI \rightarrow AgI_{(s)} + 2K_2S_2O_3 + 2NaBr \) Based on these reactions, we concluded that the solubility product constants (\(K_{sp}\)) for the silver halides follow this order: \(K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)\).

Step 1: Formation of Silver Chloride Precipitate

When sodium chloride is added to a solution of silver nitrate, a white precipitate of silver chloride is formed. The balanced chemical equation for this reaction is: \(AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3\)

Step 2: Dissolution of Silver Chloride Precipitate in Ammonia

When ammonia is added to the mixture containing the silver chloride precipitate, the precipitate dissolves. This occurs because ammonia forms a complex ion with silver ions. The balanced chemical equation for this reaction is: \(AgCl_{(s)} + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\)

Step 3: Formation of Silver Bromide Precipitate

When potassium bromide is added to the solution, a pale yellow precipitate of silver bromide appears. The balanced chemical equation for this reaction is: \[ [Ag(NH_3)_2]^+ + Cl^- + KBr \rightarrow AgBr_{(s)} + 2NH_3 + KCl\]

Step 4: Dissolution of Silver Bromide Precipitate in Sodium Thiosulfate

When a solution of sodium thiosulfate is added, the yellow silver bromide precipitate dissolves. The balanced chemical equation for this reaction is: \(AgBr_{(s)} + 2Na_2S_2O_3 \rightarrow [Ag(S_2O_3)_2]^{3-} + 2NaBr\)

Step 5: Formation of Silver Iodide Precipitate

Finally, when potassium iodide is added to the solution, a yellow precipitate of silver iodide forms. The balanced chemical equation for this reaction is: \[ [Ag(S_2O_3)_2]^{3-} + 2NaBr + 2KI \rightarrow AgI_{(s)} + 2K_2S_2O_3 + 2NaBr \]

Step 6: Final Analysis: Comparing \(K_{sp}\) Values

As we observed, AgCl, AgBr, and AgI form precipitates under different conditions, and their solubility varies in the presence of various complexing agents. This gives us an idea about the relative solubility product constants (\(K_{sp}\)) for each silver halide. Since AgCl precipitated first and then dissolved when ammonia was added, we can infer that its \(K_{sp}\) is the highest among these three silver halides. AgBr precipitated after the addition of ammonia and dissolved in sodium thiosulfate, indicating that its \(K_{sp}\) is lower than that of AgCl but higher than that of AgI. AgI precipitated last and did not dissolve in any of the complexing agents, suggesting that its \(K_{sp}\) is the lowest among the three. In conclusion, we can arrange the silver halides in the order of their decreasing solubility product constant values as follows: \(K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)\)

Key Concepts

Chemical Equations

Writing chemical equations is like crafting the perfect recipe in chemistry. These equations help express what happens during a chemical reaction. They show the reactants, which are the starting substances, and the products, the new substances formed after the reaction. Let's break this down with an example from the original exercise.

When sodium chloride (\(NaCl\)) is added to silver nitrate (\(AgNO_3\)), a white precipitate of silver chloride (\(AgCl\)) forms. The chemical equation is:

• Reactants: \(AgNO_3\), \(NaCl\)
• Products: \(AgCl_{(s)}\), \(NaNO_3\)

This is written as: \(AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3\). The subscript \((s)\) indicates that \(AgCl\) is a solid, showing its formation as a precipitate.

Always remember, balancing these equations is key. It ensures the same number of each type of atom appears on both sides, maintaining the conservation of mass. This exercise clearly demonstrates the importance of chemical equations in illustrating transformations in a reaction.

Precipitation Reactions

Precipitation reactions are fascinating processes in chemistry where two solutions mix to form an insoluble solid, known as a precipitate. In the exercise, as you added sodium chloride to a solution of silver nitrate, a white precipitate, silver chloride (\(AgCl\)), is formed. This reaction provides a wonderful example of a precipitation reaction.

How does this happen? When ionic compounds dissolve in water, they dissociate into ions. If the product of the reaction between these ions is insoluble, they clump together to form a solid precipitate. This solid is easily recognizable because it creates a cloudy or opaque appearance in the solution.

• Silver chloride (\(AgCl\)) forms a white precipitate.
• Silver bromide (\(AgBr\)), a pale yellow precipitate.
• Silver iodide (\(AgI\)), a yellow precipitate.

These formations indicate the limited solubility of these compounds, and their solubility products (\(K_{sp}\)) are essential in predicting whether a precipitate will form when two solutions are mixed.

Complex Ions Formation

Complex ions form when simple ions in solution interact with molecules or ions, often leading to visually striking changes in the solution. For example, in the exercise, adding ammonia (\(NH_3\)) to a silver chloride precipitate led to the formation of a complex ion. This dissolved the precipitate. The silver ions (\(Ag^+\)) combined with ammonia to form a complex ion \([Ag(NH_3)_2]^+\).

Complex ions play a key role in solubility. They can greatly increase the solubility of certain precipitates in a solution by transforming them into different soluble compounds.

• Ammonia complex with silver: \(AgCl_{(s)} + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\)
• Thiosulfate complex with silver: \(AgBr_{(s)} + 2Na_2S_2O_3 \rightarrow [Ag(S_2O_3)_2]^{3-} + 2NaBr\)

So, in many cases, even if an initial precipitate forms, subsequent reactions with complexing agents can make it dissolve again.

Silver Halides

Silver halides, a class of chemical compounds, involve silver paired with halogens like chlorine, bromine, and iodine. In the exercise, you encountered silver chloride (\(AgCl\)), silver bromide (\(AgBr\)), and silver iodide (\(AgI\)). These compounds commonly form precipitates due to their low solubility in water.

Their solubility products (\(K_{sp}\)) are small values indicating how much of the compound can dissolve in a solution before it precipitates out as a solid. It goes like this:

• \(AgCl\): Most soluble of the silver halides, forming white precipitate.
• \(AgBr\): Less soluble than \(AgCl\), forms a pale yellow precipitate.
• \(AgI\): Least soluble, forms a yellow precipitate not dissolvable by ammonia or thiosulfate.

Thus, the order of solubility from most to least is \(AgCl, AgBr,\) and \(AgI\). This order is mirrored by their solubility product constants, and helps predict whether a silver halide will precipitate in a particular reaction.