Problem 70 - Chemistry

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 series of reactions, the following equations correspond to the observed changes: 1. \(AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)\) 2. \(AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)\) 3. \([Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq) + KBr(aq) \rightarrow AgBr(s) + 2NH_{3}(aq) + KCl(aq)\) 4. \(AgBr(s) + 2Na_{2}S_{2}O_{3}(aq) \rightarrow [Ag(S_{2}O_{3})_{2}]^{3-}(aq) + 2NaBr(aq)\) 5. \([Ag(S_{2}O_{3})_{2}]^{3-}(aq) + 2NaBr(aq) + KI(aq) \rightarrow AgI(s) + 2Na_{2}S_{2}O_{3}(aq) + KBr(aq)\) Based on these reactions and the solubility of the silver compounds in the presence of ammonia and sodium thiosulfate, we can conclude that: \(K_{\mathrm{sp}} (\mathrm{AgCl}) > K_{\mathrm{sp}} (\mathrm{AgBr}) > K_{\mathrm{sp}} (\mathrm{AgI})\) This means silver chloride has the highest solubility, while silver iodide has the lowest solubility and the highest stability as a precipitate.

Step 1: 1. Formation of silver chloride precipitate

When sodium chloride (NaCl) solution is added to silver nitrate (AgNO3) solution, a white precipitate of silver chloride (AgCl) is formed. The balanced chemical equation for this double-displacement reaction is: \(AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)\)

Step 2: 2. Dissolution of AgCl precipitate in ammonia

When ammonia (NH3) is added to the mixture, the AgCl precipitate dissolves. This is because ammonia forms a water-soluble complex ion with silver, as shown in the following equation: \(AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)\)

Step 3: 3. Formation of silver bromide precipitate

Adding potassium bromide (KBr) solution to the mixture causes the pale yellow precipitation of silver bromide (AgBr). The balanced chemical equation for this reaction is: \([Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq) + KBr(aq) \rightarrow AgBr(s) + 2NH_{3}(aq) + KCl(aq)\)

Step 4: 4. Dissolution of AgBr precipitate in sodium thiosulfate

The yellow AgBr precipitate dissolves when sodium thiosulfate (Na2S2O3) is added, forming a soluble complex with silver: \(AgBr(s) + 2Na_{2}S_{2}O_{3}(aq) \rightarrow [Ag(S_{2}O_{3})_{2}]^{3-}(aq) + 2NaBr(aq)\)

Step 5: 5. Formation of silver iodide precipitate

Finally, adding potassium iodide (KI) to the solution results in the precipitation of yellow silver iodide (AgI). The balanced chemical equation for this reaction is: \([Ag(S_{2}O_{3})_{2}]^{3-}(aq) + 2NaBr(aq) + KI(aq) \rightarrow AgI(s) + 2Na_{2}S_{2}O_{3}(aq) + KBr(aq)\) From these reactions, we observe that \(AgCl\) dissolves in the presence of ammonia (and remains dissolved when KBr is introduced), while \(AgBr\) dissolves in the presence of sodium thiosulfate, but \(AgI\) remains precipitated throughout the entire process. This allows us to draw conclusions about the solubility product constants for these silver compounds.

Step 6: 6. Comparing K_sp values

The solubility product constant (K_sp) describes the equilibrium between a solid ionic compound and their dissolved ions in a saturated solution. A smaller solubility product constant indicates lower solubility and a more stable solid precipitate. Based on the experiment observations: \(K_{\mathrm{sp}} (\mathrm{AgCl}) > K_{\mathrm{sp}} (\mathrm{AgBr}) > K_{\mathrm{sp}} (\mathrm{AgI})\) This means that silver chloride has the highest solubility (lesser stability as a precipitate) among the three silver compounds, while silver iodide has the lowest solubility and the highest stability as a precipitate.

Key Concepts

Precipitation Reactions

Precipitation reactions take place when two aqueous solutions are mixed, resulting in the formation of an insoluble solid called a precipitate. In the given exercise, when sodium chloride (NaCl) is added to silver nitrate (AgNO_3), a white precipitate of silver chloride (AgCl) is formed.
The formation of this solid can be represented by the balanced chemical equation:

• \(AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)\)

The ions from the two solutions exchange partners, creating the new compound, AgCl, which is not soluble in water. The appearance of the precipitate indicates that equilibrium has shifted towards the formation of the solid compound.
This mechanism is controlled by solubility rules that guide which combinations of ions will result in precipitation.

Complex Ion Formation

Complex ion formation involves the binding of simple ions with molecules or other ions to form a complex ion, which is generally more soluble. In our observed reactions, ammonia (NH_3) plays a crucial role in dissolving the AgCl precipitate.
This occurs due to the formation of a complex ion, \([Ag(NH_3)_2]^+\), where the silver ion bonds with two ammonia molecules, becoming water-soluble:

• \(AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)\)

The complex formation increases the solubility of silver compounds.
Later, in the reaction where sodium thiosulfate is added, a different kind of complex ion \([Ag(S_2O_3)_2]^{3-}\) is formed, allowing the dissolution of AgBr.
These complex ions are critical in driving equilibrium towards ion dissolution, overriding typical solubility rules.

Solubility Rules

Solubility rules offer guidelines that predict whether a precipitation reaction will occur when solutions are mixed. These rules are essential for understanding which combinations of ions will result in the formation of a precipitate.
For instance, chloride salts are generally soluble, yet silver chloride (AgCl), silver bromide (AgBr), and silver iodide (AgI) are exceptions.

• AgCl forms when NaCl and AgNO_3 are combined, illustrating an exception where a commonly insoluble compound precipitates out of solution.

Understanding these exceptions helps predict and explain why certain ions come together to form a solid, instead of remaining in solution.
Each combination has solubility limitations, leading to the observed precipitation reactions, directed by the tendency of ions to form a more stable compound.

Equilibrium in Chemical Reactions

Equilibrium in chemical reactions represents the state where the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time. In the case of solubility product constants (K_sp), this principle governs the dissolution and precipitation process.
The K_sp determines the solubility of a compound: lower K_sp implies less solubility and greater precipitation.

• A high K_sp value for AgCl indicates relatively high solubility, so it dissolves in the presence of ammonia.
• In contrast, AgI precipitates even as other reactants, like sodium thiosulfate, are introduced, signifying a very low K_sp value.

The equilibrium shifts depending on the conditions and the presence of other compounds. Ammonia and sodium thiosulfate affect this balance by forming complex ions, making some reactions reversible, depending on the chemical environment.
These ideas are crucial for predicting the products and outcomes, based on differing solubility equilibria and reaction conditions.