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Consider the total ionic equation below:

[tex]\[ 2 H^{+} + CrO_4^{2-} + Ba^{2+} + 2 OH^{-} \longrightarrow Ba^{2+} + CrO_4^{2-} + 2 H_2O \][/tex]

What are the spectator ions in this equation?

A. [tex]\(CrO_4^{2-}\)[/tex] and [tex]\(OH^{-}\)[/tex]

B. [tex]\(CrO_4^{2-}\)[/tex] and [tex]\(Ba^{2+}\)[/tex]

Sagot :

GinnyAnswer
To determine the spectator ions in a given chemical equation, we first need to identify the ions that remain unchanged during the reaction process. Spectator ions are ions that do not participate directly in the formation of the products and remain in solution in the same form on both sides of the equation.

Let's consider the given total ionic equation:
[tex]\[ 2 H^+ + CrO_4^{2-} + Ba^{2+} + 2 OH^- \longrightarrow Ba^{2+} + CrO_4^{2-} + 2 H_2O \][/tex]

We can now compare the reactants and products to see which ions appear on both sides of the equation without undergoing any change.

1. Reactants:
- [tex]\(2 H^+\)[/tex]
- [tex]\(CrO_4^{2-}\)[/tex]
- [tex]\(Ba^{2+}\)[/tex]
- [tex]\(2 OH^-\)[/tex]

2. Products:
- [tex]\(Ba^{2+}\)[/tex]
- [tex]\(CrO_4^{2-}\)[/tex]
- [tex]\(2 H_2O\)[/tex]

By comparing the reactant side with the product side, we observe the following:

- [tex]\(Ba^{2+}\)[/tex] appears on both sides of the equation as a reactant and a product. It remains unchanged.
- [tex]\(CrO_4^{2-}\)[/tex] also appears on both sides of the equation and remains unchanged.

The other ions, [tex]\(2 H^+\)[/tex] and [tex]\(2 OH^-\)[/tex], do not appear on both sides unchanged; instead, they react to form [tex]\(2 H_2O\)[/tex].

These unchanged ions are what we call the spectator ions. Therefore, the spectator ions are:

- [tex]\(Ba^{2+}\)[/tex]
- [tex]\(CrO_4^{2-}\)[/tex]

Conclusively, the spectator ions in the given equation are [tex]\(Ba^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex].
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