Addition of HCl to an aqueous solution of `Pb(NO_3)_2` gives a

Explanation

To solve the question regarding the addition of HCl to an aqueous solution of Pb(NO₃)₂, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are lead(II) nitrate (Pb(NO₃)₂) and hydrochloric acid (HCl).

### Step 2: Write the Chemical Equation
When lead(II) nitrate reacts with hydrochloric acid, the following reaction occurs:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{HCl} (aq) \rightarrow \text{PbCl}_2 (s) + 2 \text{HNO}_3 (aq) \]

### Step 3: Identify the Products
From the reaction, we can see that the products formed are lead(II) chloride (PbCl₂) and nitric acid (HNO₃).

### Step 4: Determine the State of the Products
- Lead(II) chloride (PbCl₂) is a solid that precipitates out of the solution, forming a white precipitate.
- Nitric acid (HNO₃) remains in the aqueous state.

### Step 5: Conclusion
The addition of HCl to an aqueous solution of Pb(NO₃)₂ results in the formation of a white precipitate of lead(II) chloride (PbCl₂).

### Final Answer
The addition of HCl to an aqueous solution of Pb(NO₃)₂ gives a white precipitate of PbCl₂.

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