`PbCl_(2)darr+"hot water"toPb^(2)(aq.)+2Cl^(-)(aq.)`

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:

\[ \text{PbCl}_2 (s) + \text{hot water} \rightarrow \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq) \]

we can follow these steps:

### Step 1: Identify the Reactants and Products
- **Reactant**: Lead(II) chloride, \( \text{PbCl}_2 \), is a solid precipitate (indicated by (s)).
- **Products**: Lead ions \( \text{Pb}^{2+} \) and chloride ions \( \text{Cl}^- \) are in aqueous solution (indicated by (aq)).

### Step 2: Analyze the State of Reactants and Products
- The reactant \( \text{PbCl}_2 \) is a precipitate, which means it is not soluble in water.
- The products, \( \text{Pb}^{2+} \) and \( 2\text{Cl}^- \), are ions that are dissolved in water.

### Step 3: Determine the Type of Reaction
- Since \( \text{PbCl}_2 \) is dissolving in hot water to form ions in solution, this indicates that the solid precipitate is breaking down into its ionic components.
- This process is known as **dissolution**, where a solid precipitate dissolves in a solvent to form a solution.

### Conclusion
The reaction is classified as a **precipitate dissolution reaction** because a solid precipitate (PbCl2) is dissolving in water to form aqueous ions (Pb2+ and Cl-).

### Final Answer
The type of reaction is: **Precipitate Dissolution Reaction**.

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