Problem 45 - Chemistry: Principles and Reactions

Question

A solution of potassium dichromate is made basic with sodium hydroxide; the color changes from red to yellow. Addition of silver nitrate to the yellow solution gives a precipitate. This precipitate dissolves in concentrated ammonia but re-forms when nitric acid is added. Write balanced net ionic equations for all the reactions in this sequence.

Step-by-Step Solution

Answer

Question: Write down the balanced net ionic equations for each step of the following reactions: 1) Potassium dichromate solution turning basic with sodium hydroxide; 2) Adding silver nitrate to the yellow solution to form a precipitate; 3) Dissolving the precipitate in concentrated ammonia; 4) Re-forming the precipitate by adding nitric acid. Answer: 1) Cr2O7^2- + 2OH^- → 2CrO4^2- 2) 2Ag^+ + CrO4^2- → Ag2CrO4(s) 3) Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2- 4) 2Ag(NH3)2^+ + CrO4^2- + 4H^+ → Ag2CrO4(s) + 4NH4^+

Step 1: Write the chemical formulas of the compounds involved in the reactions

We will determine the chemical formulas of each compound mentioned in the exercise. Potassium dichromate: K2Cr2O7 Sodium hydroxide: NaOH Silver nitrate: AgNO3 Ammonia: NH3 Nitric acid: HNO3 Now we can analyze the reactions and write the balanced net ionic equations.

Step 2: Formation of basic potassium dichromate solution

In this step, potassium dichromate (K2Cr2O7) reacts with sodium hydroxide (NaOH) to form a basic solution with a yellow color. The balanced chemical equation for this reaction is: K2Cr2O7 + 2NaOH → 2KOH + Na2CrO4 The net ionic equation is obtained by eliminating the spectator ions, which are potassium (K+) and sodium (Na+) ions, as they remain unchanged during the reaction. Net ionic equation: Cr2O7^2- + 2OH^- → 2CrO4^2-

Step 3: Formation of a precipitate by adding silver nitrate

In this reaction, silver nitrate (AgNO3) is added to the yellow solution from step 2. The silver ions (Ag+) react with the chromate ions (CrO4^2-) to form a precipitate, silver chromate (Ag2CrO4). AgNO3 + Na2CrO4 → Ag2CrO4(s) + 2NaNO3 Cancel out the spectator ions (Na+ and NO3^-): Net ionic equation: 2Ag^+ + CrO4^2- → Ag2CrO4(s)

Step 4: Precipitate dissolving in concentrated ammonia

Silver chromate (Ag2CrO4) precipitate will dissolve in ammonia (NH3) to form a complex ion. Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2- Since there are no spectator ions, the net ionic equation is the same as the balanced chemical equation. Net ionic equation: Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2-

Step 5: Precipitate re-forming when nitric acid is added

In this final step, nitric acid (HNO3) is added to the solution, causing the silver complex ion to decompose and reform the silver chromate (Ag2CrO4) precipitate. 2Ag(NH3)2^+ + CrO4^2- + 4HNO3 → Ag2CrO4(s) + 4NH4NO3 Eliminate the spectator ions (NO3^-): Net ionic equation: 2Ag(NH3)2^+ + CrO4^2- + 4H^+ → Ag2CrO4(s) + 4NH4^+ These are the balanced net ionic equations for each reaction in this sequence.

Key Concepts

Potassium Dichromate Reaction

In this reaction, potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) is mixed with sodium hydroxide (\( \text{NaOH} \)) to create a basic solution. Initially, potassium dichromate is known for its distinctive red color. When it reacts with a base like sodium hydroxide, the chromate ions (\( \text{Cr}_2\text{O}_7^{2-} \)) convert into chromate ions (\( \text{CrO}_4^{2-} \)), resulting in a color change from red to yellow. This transformation indicates the formation of chromate ions.

To balance the chemical equation, we need to consider that sodium and potassium ions do not change during the reaction, and are therefore eliminated in the net ionic equation. This simplifies to:

• Initial Ionic Equation: \( \text{Cr}_2\text{O}_7^{2-} + 2\text{NaOH} \rightarrow 2\text{NaCrO}_4 + \text{H}_2\text{O} \)
• Net Ionic Equation: \( \text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} \)

This fundamental reaction showcases how altering the pH can result in discernible changes, such as color variations.

Precipitation Reactions

Precipitation reactions occur when two soluble ions in solution combine to form an insoluble compound, or precipitate, which then falls out of solution as a solid. In this example, adding silver nitrate (\( \text{AgNO}_3 \)) to the yellow chromate solution results in the formation of silver chromate (\( \text{Ag}_2\text{CrO}_4 \)), a red-brown precipitate. This is a classic example of a precipitation reaction.

The silver ions (\( \text{Ag}^+ \)) and the chromate ions (\( \text{CrO}_4^{2-} \)) interact in the solution to form an insoluble compound.

• Initial Ionic Equation: \( 2\text{AgNO}_3 + \text{Na}_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2\text{NaNO}_3 \)
• Net Ionic Equation: \( 2\text{Ag}^+ + \text{CrO}_4^{2-} \rightarrow \text{Ag}_2\text{CrO}_4(s) \)

Understanding precipitation reactions is crucial in predicting and visually confirming chemical reactions, as the formation of a solid precipitate is often a clear indicator of a reaction.

Complex Ion Formation

A complex ion is a chemical structure consisting of a central metal ion bonded to surrounding molecules or ions. In this reaction sequence, the silver chromate (\( \text{Ag}_2\text{CrO}_4 \)) precipitate dissolves upon the addition of ammonia (\( \text{NH}_3 \)), forming a complex ion. The ammonia molecules coordinate with the silver ions (\( \text{Ag}^+ \)) to produce the silver diamine complex, \( \text{Ag(NH}_3)_2^+ \).

This reaction illustrates a shift from a simple precipitate to a soluble complex ion.

• Initial Ionic Equation: \( \text{Ag}_2\text{CrO}_4(s) + 4\text{NH}_3 \rightarrow 2\text{Ag(NH}_3)_2^+ + \text{CrO}_4^{2-} \)
• Net Ionic Equation: Identical to the initial, as no additional spectator ions need to be accounted for.

This shows the capacity of ammonia to stabilize and dissolve certain precipitates by forming soluble ionic complexes, which is useful in various chemical dissolution processes.

Silver Chromate Precipitation

The silver chromate (\( \text{Ag}_2\text{CrO}_4 \)) precipitation reappears when nitric acid (\( \text{HNO}_3 \)) is introduced, reversing the dissolving process. The item that was dissolved as a complex ion reforms back into an insoluble solid, as the \( \text{Ag(NH}_3)_2^+ \) is decomposed by the acid.

As the ammonia complexes are broken down by the nitric acid, the original silver chromate precipitate reforms.

• Initial Ionic Equation: \( 2\text{Ag(NH}_3)_2^+ + \text{CrO}_4^{2-} + 4\text{HNO}_3 \rightarrow \text{Ag}_2\text{CrO}_4(s) + 4\text{NH}_4\text{NO}_3 \)
• Net Ionic Equation: \( 2\text{Ag(NH}_3)_2^+ + \text{CrO}_4^{2-} + 4\text{H}^+ \rightarrow \text{Ag}_2\text{CrO}_4(s) + 4\text{NH}_4^+ \)

This exemplifies how the addition of an acid can alter the solubility of complex ions, leading to the re-precipitation of insoluble solids.