Problem 34 - Chemistry The Molecular Science

Question

If aqueous solutions of potassium sulfide and iron(III) chloride are mixed, a precipitate is formed. Write the complete and net ionic equations for this reaction, and name the precipitate.

Step-by-Step Solution

Answer

The net ionic equation is \( 3S^{2-} + 2Fe^{3+} \rightarrow Fe_2S_3 \); the precipitate is iron(III) sulfide.

Step 1: Balancing the Chemical Equation

Start by writing the balanced chemical equation for the reaction. Potassium sulfide (\( K_2S \)) reacts with iron(III) chloride (\( FeCl_3 \)) to form iron(III) sulfide (\( Fe_2S_3 \)), which is the precipitate, and potassium chloride (\( KCl \)): \[ 3K_2S (aq) + 2FeCl_3 (aq) \rightarrow Fe_2S_3 (s) + 6KCl (aq) \].

Step 2: Dissociating Species Into Ions

Write each aqueous compound as its constituent ions. Potassium sulfide dissociates into \( 2K^+ \) and \( S^{2-} \); iron(III) chloride dissociates into \( 2Fe^{3+} \) and \( 3Cl^- \): \[ 3(2K^+ + S^{2-}) + 2(Fe^{3+} + 3Cl^-) \].

Step 3: Writing the Complete Ionic Equation

Include all ions present in the reaction. The complete ionic equation becomes:\[ 6K^+ + 3S^{2-} + 2Fe^{3+} + 6Cl^- \rightarrow Fe_2S_3(s) + 6K^+ + 6Cl^- \].

Step 4: Identifying Spectator Ions

Identify and cancel out the spectator ions, which are the ions that appear unchanged on both sides of the equation. The spectator ions in this reaction are \( K^+ \) and \( Cl^- \).

Step 5: Writing the Net Ionic Equation

After removing the spectator ions, write the net ionic equation for the formation of the precipitate:\[ 3S^{2-}(aq) + 2Fe^{3+}(aq) \rightarrow Fe_2S_3(s) \].

Step 6: Naming the Precipitate

The solid formed is iron(III) sulfide, represented by the chemical formula \( Fe_2S_3 \). This is the precipitate in the reaction.

Key Concepts

Chemical Reaction

When we talk about a chemical reaction, we are referring to a process in which substances (the reactants) change into new substances (the products). This transformation occurs through the breaking and forming of chemical bonds. In the exercise provided, the chemical reaction involves potassium sulfide reacting with iron(III) chloride.

• Reactants: Potassium sulfide \(K_2S\) and iron(III) chloride \(FeCl_3\)
• Products: Iron(III) sulfide \(Fe_2S_3\) and potassium chloride \(KCl\)

At the molecular level, the compounds react in a way that the ions switch partners. Potassium ions team up with chloride ions, and sulfide ions pair with iron ions. This is expressed in the reaction: \[3K_2S (aq) + 2FeCl_3 (aq) \rightarrow Fe_2S_3 (s) + 6KCl (aq)\]The balanced equation confirms that mass and charge are conserved during the reaction.

Ionic Equation

The ionic equation provides a clearer picture of what happens on the microscopic level. It breaks down aqueous compounds into their respective ions.In the provided example, the dissociation process can be shown as follows:

• Potassium sulfide: \(3K_2S (aq) \rightarrow 6K^+ (aq) + 3S^{2-} (aq)\)
• Iron(III) chloride: \(2FeCl_3 (aq) \rightarrow 2Fe^{3+} (aq) + 6Cl^- (aq)\)

Writing the complete ionic equation involves showing all ions present in a chemical reaction. This method helps identify the species involved directly in chemical changes versus those that remain unchanged. Hence, the complete ionic form of the equation appears as:\[ 6K^+ + 3S^{2-} + 2Fe^{3+} + 6Cl^- \rightarrow Fe_2S_3(s) + 6K^+ + 6Cl^- \]This equation serves as a foundation for simplifying reactions further into net ionic form.

Precipitate Formation

Precipitate formation occurs when two aqueous solutions react and form an insoluble solid. This solid, called a precipitate, emerges from the solution as a distinct phase.In the exercise, iron(III) sulfide \(Fe_2S_3\) forms as a precipitate. This compound is insoluble in water, causing it to separate from the solution upon mixing the reactants.Why do precipitates form?- Insolubility: Certain ions, when combined, form compounds that don't dissolve in water.- Phase separation: The solid separates out because its formation is more stable energetically in the solid state.By observing precipitate formation, we gain insights into solubility rules and the interaction between ions in solution.

Spectator Ions

Spectator ions are ions that exist in the same form on both the reactant and product side of a chemical equation. They do not participate in the chemical change but simply "watch" the reaction unfold.In the iron(III) sulfide formation, the spectator ions are:

• Potassium ions \(K^+\)
• Chloride ions \(Cl^-\)

These ions maintain their presence in the solution unchanged, as demonstrated in the complete ionic equation:\[ 6K^+ + 3S^{2-} + 2Fe^{3+} + 6Cl^- \rightarrow Fe_2S_3(s) + 6K^+ + 6Cl^- \]Upon removing these spectator ions, we arrive at the net ionic equation:\[ 3S^{2-}(aq) + 2Fe^{3+}(aq) \rightarrow Fe_2S_3(s) \]Understanding spectator ions helps streamline chemical equations by focusing on actual chemical changes.