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What is the total ionic equation for the following reaction?

[tex]\[ H_2CrO_4 + Ba(OH)_2 \][/tex]

A. [tex]\[ H_2CrO_4^{-} + 2 OH^{-} \rightarrow CrO_4^{-} + 2 H_2O \][/tex]

B. [tex]\[ 2 H^{+} + CrO_4^{-} + 2 OH^{-} \rightarrow CrO_4^{-} + 2 H_2O \][/tex]

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


Sagot :

To determine the total ionic equation, we carefully observe the dissociation of each substance in aqueous solution and write out all the ions involved in the reaction. Here's a step-by-step breakdown of the process:

1. Dissociation of Reactants in Aqueous Solution:

- Chromic acid ([tex]\( H_2CrO_4 \)[/tex]):
[tex]\[ H_2CrO_4 \rightarrow 2H^+ + CrO_4^{2-} \][/tex]

- Barium hydroxide ([tex]\( Ba(OH)_2 \)[/tex]):
[tex]\[ Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^- \][/tex]

2. Total Ionic Equation:
Combine all the dissociated ions from the above reactants on the left side of the equation:
[tex]\[ 2H^+ + CrO_4^{2-} + Ba^{2+} + 2OH^- \rightarrow \][/tex]

3. Formation of Products in Aqueous Solution:

- Barium chromate ([tex]\( BaCrO_4 \)[/tex]) is an insoluble salt and remains undissociated in water, but in this case it is not formed as a product in the given reaction.

- Water ([tex]\( H_2O \)[/tex]) is a molecular compound that does not dissociate in solution.

Therefore, the product side of the equation remains the ions [tex]\( Ba^{2+} \)[/tex] and [tex]\( CrO_4^{2-} \)[/tex] along with the formed water:
[tex]\[ Ba^{2+} + CrO_4^{2-} + 2H_2O \][/tex]

4. Complete Total Ionic Equation:
Putting these together, the total ionic equation is:
[tex]\[ 2H^+ + CrO_4^{2-} + Ba^{2+} + 2OH^- \rightarrow Ba^{2+} + CrO_4^{2-} + 2H_2O \][/tex]

This final equation represents the total ionic equation of the given reactions involving chromic acid and barium hydroxide in aqueous solution.
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