Answer :
Certainly! To identify the spectator ions in the given reaction, we need to carefully analyze the reactants and products to see which ions remain unchanged throughout the reaction.
The given reaction is:
[tex]\[ H^+ + Br^- + K^+ + OH^- \rightarrow K^+ + Br^- + H_2O \][/tex]
Here are the individual ions and molecules involved:
1. Reactants: [tex]\( H^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( K^+ \)[/tex], [tex]\( OH^- \)[/tex]
2. Products: [tex]\( K^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( H_2O \)[/tex]
### Step-by-Step Solution
1. Identify the reactants and products:
- On the reactant side: [tex]\( H^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( K^+ \)[/tex], [tex]\( OH^- \)[/tex]
- On the product side: [tex]\( K^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( H_2O \)[/tex]
2. Identify what changes and what remains the same:
- [tex]\( H^+ \)[/tex] (Hydrogen ion) and [tex]\( OH^- \)[/tex] (Hydroxide ion) react to form water [tex]\( H_2O \)[/tex].
- The ions [tex]\( K^+ \)[/tex] (Potassium ion) and [tex]\( Br^- \)[/tex] (Bromide ion) remain unchanged on both sides of the equation.
3. Determine the spectator ions:
- Spectator ions are those that do not participate in the chemical reaction and remain in the ionic form both before and after the reaction.
4. Conclusion:
- The ions [tex]\( K^+ \)[/tex] and [tex]\( Br^- \)[/tex] do not participate in the formation of [tex]\( H_2O \)[/tex] and remain unchanged.
Thus, the spectator ions in the reaction are [tex]\( K^+ \)[/tex] and [tex]\( Br^- \)[/tex].
So, the answer is:
[tex]\[ \text{K}^{+} \text{and} \text{ Br}^{-} \][/tex]
This means that [tex]\( K^{+} \)[/tex] and [tex]\( Br^{-} \)[/tex] are the ions that do not change during the course of the reaction and are therefore the spectator ions.
The given reaction is:
[tex]\[ H^+ + Br^- + K^+ + OH^- \rightarrow K^+ + Br^- + H_2O \][/tex]
Here are the individual ions and molecules involved:
1. Reactants: [tex]\( H^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( K^+ \)[/tex], [tex]\( OH^- \)[/tex]
2. Products: [tex]\( K^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( H_2O \)[/tex]
### Step-by-Step Solution
1. Identify the reactants and products:
- On the reactant side: [tex]\( H^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( K^+ \)[/tex], [tex]\( OH^- \)[/tex]
- On the product side: [tex]\( K^+ \)[/tex], [tex]\( Br^- \)[/tex], [tex]\( H_2O \)[/tex]
2. Identify what changes and what remains the same:
- [tex]\( H^+ \)[/tex] (Hydrogen ion) and [tex]\( OH^- \)[/tex] (Hydroxide ion) react to form water [tex]\( H_2O \)[/tex].
- The ions [tex]\( K^+ \)[/tex] (Potassium ion) and [tex]\( Br^- \)[/tex] (Bromide ion) remain unchanged on both sides of the equation.
3. Determine the spectator ions:
- Spectator ions are those that do not participate in the chemical reaction and remain in the ionic form both before and after the reaction.
4. Conclusion:
- The ions [tex]\( K^+ \)[/tex] and [tex]\( Br^- \)[/tex] do not participate in the formation of [tex]\( H_2O \)[/tex] and remain unchanged.
Thus, the spectator ions in the reaction are [tex]\( K^+ \)[/tex] and [tex]\( Br^- \)[/tex].
So, the answer is:
[tex]\[ \text{K}^{+} \text{and} \text{ Br}^{-} \][/tex]
This means that [tex]\( K^{+} \)[/tex] and [tex]\( Br^{-} \)[/tex] are the ions that do not change during the course of the reaction and are therefore the spectator ions.
