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

Options

A. For precipitate 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 \( \text{PbCrO}_4 + 4 \text{NaOH (excess)} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] \), we can follow these steps:

### Step 1: Identify the Reactants and Products
- The reactants are lead chromate (\( \text{PbCrO}_4 \)) and sodium hydroxide (\( \text{NaOH} \)).
- The product is sodium lead hydroxide (\( \text{Na}_2[\text{Pb(OH)}_4] \)).

### Step 2: Analyze the State of the Reactants
- Lead chromate (\( \text{PbCrO}_4 \)) is a yellow precipitate, indicating it is insoluble in water.
- Sodium hydroxide (\( \text{NaOH} \)) is a strong base and is soluble in water.

### Step 3: Determine the Type of Reaction
- The reaction involves the dissolution of a precipitate (\( \text{PbCrO}_4 \)) in excess sodium hydroxide.
- Since the precipitate is present in the reactants and is being converted into a soluble complex (\( \text{Na}_2[\text{Pb(OH)}_4] \)), this indicates that the precipitate is dissolving.

### Step 4: Classify the Reaction
- Given that the precipitate is dissolving in the presence of excess sodium hydroxide, this reaction is classified as a **precipitate dissolution reaction**.

### Conclusion
The type of reaction for the given equation is a **precipitate dissolution reaction**.

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