Consider the total ionic equation below.

[tex]\[
2 \text{NH}_4^{+} + 2 \text{OH}^{-} + 2 \text{H}^{+} + \text{SO}_4^{2-} \rightarrow 2 \text{NH}_4^{+} + 2 \text{H}_2 \text{O} + \text{SO}_4^{2-}
\][/tex]

What are the spectator ions in this equation?

A. [tex]\(\text{H}^{+}\)[/tex] and [tex]\(\text{OH}^{-}\)[/tex]

B. [tex]\(\text{H}^{+}\)[/tex] and [tex]\(\text{NH}_4^{+}\)[/tex]

C. [tex]\(\text{NH}_4^{+}\)[/tex] and [tex]\(\text{SO}_4^{2-}\)[/tex]

D. [tex]\(\text{SO}_4^{2-}\)[/tex] and [tex]\(\text{OH}^{-}\)[/tex]



Answer :

To identify the spectator ions in the total ionic equation, we need to look at which ions remain unchanged on both sides of the equation. Spectator ions are those that do not participate in the chemical reaction but remain in the solution unchanged.

Let's examine the given total ionic equation:
[tex]\[ 2 NH_4^+ + 2 OH^- + 2 H^+ + SO_4^{2-} \rightarrow 2 NH_4^+ + 2 H_2O + SO_4^{2-} \][/tex]

1. Identifying NH_4^+ (Ammonium Ion):
- On the reactant side: There are 2 [tex]\(NH_4^+\)[/tex].
- On the product side: There are also 2 [tex]\(NH_4^+\)[/tex].
- Since [tex]\(NH_4^+\)[/tex] appears unchanged on both sides, it is a spectator ion.

2. Identifying OH^- (Hydroxide Ion):
- On the reactant side: There are 2 [tex]\(OH^-\)[/tex].
- On the product side: [tex]\(OH^-\)[/tex] is not present; it has reacted to form water ([tex]\(H_2O\)[/tex]).
- Since [tex]\(OH^-\)[/tex] undergoes a change, it is not a spectator ion.

3. Identifying H^+ (Hydrogen Ion):
- On the reactant side: There are 2 [tex]\(H^+\)[/tex].
- On the product side: [tex]\(H^+\)[/tex] is not present; it has reacted to form water ([tex]\(H_2O\)[/tex]).
- Since [tex]\(H^+\)[/tex] undergoes a change, it is not a spectator ion.

4. Identifying SO_4^{2-} (Sulfate Ion):
- On the reactant side: There is 1 [tex]\(SO_4^{2-}\)[/tex].
- On the product side: There is also 1 [tex]\(SO_4^{2-}\)[/tex].
- Since [tex]\(SO_4^{2-}\)[/tex] appears unchanged on both sides, it is a spectator ion.

By observing the unchanged ions, we conclude that the spectator ions in the equation are:
- [tex]\( NH_4^+ \)[/tex] (Ammonium ion)
- [tex]\( SO_4^{2-} \)[/tex] (Sulfate ion)

Thus, the correct answer is:
[tex]\( NH_4^+ \)[/tex] and [tex]\( SO_4^{2-} \)[/tex]

So the spectator ions in this reaction are [tex]\(NH_4^+\)[/tex] and [tex]\(SO_4^{2-}\)[/tex].