`PbCrO_(4)darr+4NaOH(excess)toNa_(2)[Pb(OH)_(4)]+Na_(2)CrO_(4)`

Options

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

Answer

To determine the type of reaction represented by the equation:

\[ \text{PbCrO}_4 + 4 \text{NaOH (excess)} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4 \]

we can analyze the components involved and the changes that occur during the reaction.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants are lead(II) chromate (\( \text{PbCrO}_4 \)) and sodium hydroxide (\( \text{NaOH} \)).
- \( \text{PbCrO}_4 \) is known to be a yellow precipitate.

2. **Identify the Products:**
- The products of the reaction are sodium lead(II) hydroxide (\( \text{Na}_2[\text{Pb(OH)}_4] \)) and sodium chromate (\( \text{Na}_2\text{CrO}_4 \)).
- \( \text{Na}_2[\text{Pb(OH)}_4] \) is a soluble complex, and \( \text{Na}_2\text{CrO}_4 \) is also soluble.

3. **Analyze the Reaction:**
- In the reactants, we start with a precipitate (\( \text{PbCrO}_4 \)).
- As the reaction proceeds, the precipitate dissolves and forms soluble products.

4. **Determine the Type of Reaction:**
- Since a precipitate is present in the reactants and it dissolves to form soluble products, this indicates that the reaction involves the dissolution of a precipitate.
- Therefore, this reaction is classified as a **precipitate dissolution reaction**.

### Conclusion:
The reaction \( \text{PbCrO}_4 + 4 \text{NaOH} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4 \) is a **precipitate dissolution reaction**.

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