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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. \(\text{H}^{+}\) and \(\text{OH}^{-}\)

B. \(\text{H}^{+}\) and \(\text{NH}_4^{+}\)

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

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



Answer :

GinnyAnswer
To determine the spectator ions in the given total ionic equation, we should first identify the ions that appear unchanged on both sides of the equation. Spectator ions are those that do not participate in the actual chemical reaction but are present in the solution.

The given total ionic equation is:
[tex]\[ 2 NH_4^+ + 2 OH^- + 2 H^+ + SO_4^{2-} \rightarrow 2 NH_4^+ + 2 H_2O + SO_4^{2-} \][/tex]

Now, let's examine each ion:
- \( NH_4^+ \) (ammonium ion) appears on both the reactant side (left) and the product side (right) in an unchanged form.
- \( OH^- \) (hydroxide ion) on the reactant side reacts with \( H^+ \) (hydrogen ion) to form water \( H_2O \), so they do not remain unchanged.
- \( H^+ \) (hydrogen ion) reacts with \( OH^- \) (hydroxide ion) to form water \( H_2O \), so it also changes.
- \( SO_4^{2-} \) (sulfate ion) appears on both the reactant side (left) and the product side (right) unchanged.

Given this, the ions \( NH_4^+ \) and \( SO_4^{2-} \) are the ones that appear unaltered on both sides of the equation, hence they are the spectator ions.

Therefore, the spectator ions in the given total ionic equation are:
[tex]\[ NH_4^+ \text{ and } SO_4^{2-} \][/tex]

The correct option is:
[tex]\[ \boxed{NH_4^+ \text{ and } SO_4^{2-}} \][/tex]

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