`Pb(AcO)_(2)+Na_(2)CrO_(4) to PbCrO_(4) darr+2AcONa`

Options

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

Answer

To solve the given reaction:

**Step 1: Identify the reactants and products.**
The reactants are lead(II) acetate, \( \text{Pb(AcO)}_2 \), and sodium chromate, \( \text{Na}_2\text{CrO}_4 \). The products are lead(II) chromate, \( \text{PbCrO}_4 \), and sodium acetate, \( 2\text{AcONa} \).

**Step 2: Write the balanced chemical equation.**
The balanced equation for the reaction is:
\[ \text{Pb(AcO)}_2 + \text{Na}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 \downarrow + 2\text{AcONa} \]
Here, the downward arrow indicates that lead(II) chromate, \( \text{PbCrO}_4 \), is a precipitate.

**Step 3: Classify the type of reaction.**
This reaction can be classified as a double displacement reaction (also known as a metathesis reaction). In this type of reaction, the cations and anions of the reactants exchange partners to form new products.

**Step 4: Identify the precipitate.**
In the products, \( \text{PbCrO}_4 \) is a solid precipitate, which forms when lead(II) ions react with chromate ions in solution. The formation of a precipitate is a key indicator of a chemical reaction taking place.

**Step 5: Conclusion.**
Since a precipitate is formed in this reaction, it can be concluded that this is a precipitate formation reaction.

**Final Answer:**
The reaction is classified as a **precipitate formation reaction**.

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