`SrCrO_(4)darr+2AcOH(excess)toSr(ArO)_(2)to`No dissolution

Options

A. For precipitate formation reaction
B. For precipitate dissolution reaction
C. For precipitate exchange reaction
D. For no reaction

Answer

To solve the question regarding the reaction of strontium chromate (SrCrO4) with acetic acid (AcOH), we will go through the steps of the reaction and analyze the products formed.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is strontium chromate (SrCrO4), which is a yellow precipitate.
- The solvent is acetic acid (AcOH), which is present in excess.

2. **Understanding the Reaction**:
- When SrCrO4 is mixed with acetic acid, the acetic acid acts as a solvent and can dissolve the precipitate.
- The chemical formula for acetic acid is CH3COOH.

3. **Reaction Process**:
- The strontium chromate (SrCrO4) will dissolve in the excess acetic acid.
- This dissolution leads to the formation of a soluble complex, which is strontium acetate (Sr(CH3COO)2).

4. **Formation of Products**:
- The product formed from the reaction is strontium acetate (Sr(CH3COO)2).
- Since the acetic acid is in excess, the initial precipitate (SrCrO4) completely dissolves, and there is no precipitate remaining in the solution.

5. **Final Observation**:
- After the reaction, there is no precipitate left in the solution, indicating that the strontium chromate has completely dissolved in the acetic acid.

6. **Conclusion**:
- The type of reaction that occurs here is a **precipitate dissolution reaction** because the initial precipitate (SrCrO4) dissolves in the solvent (acetic acid), resulting in a solution with no precipitate.

### Final Answer:
The correct answer to the question is that the reaction results in **precipitate dissolution**.