To determine the spectator ions in the given total ionic equation, we need to identify which ions appear on both sides of the equation without undergoing any change.
The total ionic equation given is:
[tex]\[ 2 H^+ + CrO_4^{2-} + Ba^{2+} + 2 OH^- \longrightarrow Ba^{2+} + CrO_4^{2-} + 2 H_2O \][/tex]
Let's analyze this equation step by step:
1. On the left-hand side of the equation:
- [tex]\( 2 H^+ \)[/tex] (hydrogen ions)
- [tex]\( CrO_4^{2-} \)[/tex] (chromate ions)
- [tex]\( Ba^{2+} \)[/tex] (barium ions)
- [tex]\( 2 OH^- \)[/tex] (hydroxide ions)
2. On the right-hand side of the equation:
- [tex]\( Ba^{2+} \)[/tex] (barium ions)
- [tex]\( CrO_4^{2-} \)[/tex] (chromate ions)
- [tex]\( 2 H_2O \)[/tex] (water molecules)
Now, we compare the ions on both sides of the equation:
- [tex]\( Ba^{2+} \)[/tex] and [tex]\( CrO_4^{2-} \)[/tex] are present on both sides of the equation without undergoing any change in their form or respective states.
The ions that appear on both sides unchanged are the spectator ions. These ions do not participate directly in the chemical reaction but merely "spectate" the process.
Thus, the correct answer is:
[tex]\[ \text{CrO}_4^{2-} \text{ and } \text{Ba}^{2+} \][/tex]
So, the spectator ions in this equation are [tex]\( CrO_4^{2-} \)[/tex] and [tex]\( Ba^{2+} \)[/tex].
Hence, the correct choice is:
[tex]\[ \text{CrO}_4^{2-} \text{ and } \text{Ba}^{2+} \][/tex]
