To determine the spectator ions in the given total ionic equation, we need to identify the ions that do not undergo any change during the reaction. These ions will appear in the same form on both the reactants and products side of the equation.
The given total ionic equation is:
[tex]\[ 2H^+ (aq) + CrO_4^{2-} (aq) + Ba^{2+} (aq) + 2OH^- (aq) \longrightarrow Ba^{2+} (aq) + CrO_4^{2-} (aq) + 2H_2O (l) \][/tex]
Let's analyze each ion individually:
1. Hydrogen Ion ([tex]\(H^+\)[/tex]):
- Present on the reactant side: Yes, in the form [tex]\(2H^+\)[/tex].
- Present on the product side: No, it is involved in the formation of water ([tex]\(H_2O\)[/tex]).
2. Chromate Ion ([tex]\(CrO_4^{2-}\)[/tex]):
- Present on the reactant side: Yes, [tex]\(CrO_4^{2-}\)[/tex].
- Present on the product side: Yes, [tex]\(CrO_4^{2-}\)[/tex].
- This ion remains unchanged during the reaction.
3. Barium Ion ([tex]\(Ba^{2+}\)[/tex]):
- Present on the reactant side: Yes, [tex]\(Ba^{2+}\)[/tex].
- Present on the product side: Yes, [tex]\(Ba^{2+}\)[/tex].
- This ion remains unchanged during the reaction.
4. Hydroxide Ion ([tex]\(OH^-\)[/tex]):
- Present on the reactant side: Yes, in the form [tex]\(2OH^-\)[/tex].
- Present on the product side: No, it is involved in the formation of water ([tex]\(H_2O\)[/tex]).
From the analysis:
- The [tex]\(CrO_4^{2-}\)[/tex] ion does not change from reactants to products; therefore, it is a spectator ion.
- The [tex]\(Ba^{2+}\)[/tex] ion does not change from reactants to products; therefore, it is a spectator ion.
Thus, the spectator ions in this reaction are [tex]\( CrO_4^{2-} \)[/tex] and [tex]\( Ba^{2+} \)[/tex].
The correct answer is:
[tex]\[ \text{CrO}_4^{2-} \text{ and } \text{Ba}^{2+} \][/tex]
