A white precipitate is formed when silver nitrate is added to a solution containing chloride ions. This precipitate is soluble in 
Hydrochloric acid ; Nitric acid ; Dilute H 2 S O 4 ; Ammonium hydroxide ;

Options

A. Hydrochloric acid
B. Nitric acid
C. Dilute `H_2SO_4`
D. Ammonium hydroxide

Answer

To solve the question regarding the solubility of the white precipitate formed when silver nitrate is added to a solution containing chloride ions, we will follow these steps:

### Step 1: Identify the Reaction
When silver nitrate (AgNO₃) is added to a solution containing chloride ions (Cl⁻), a white precipitate of silver chloride (AgCl) is formed. The reaction can be represented as follows:

\[ \text{AgNO}_3 (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) + \text{NO}_3^- (aq) \]

### Step 2: Analyze the Precipitate
The precipitate formed, silver chloride (AgCl), is known to be a white solid that is only sparingly soluble in water. We need to determine in which of the given solutions this precipitate is soluble.

### Step 3: Evaluate the Options
1. **Hydrochloric acid (HCl)**: Adding HCl would not dissolve AgCl, as it would simply provide more chloride ions, which would continue to form AgCl.

2. **Nitric acid (HNO₃)**: Similar to HCl, nitric acid does not dissolve AgCl because it does not provide a mechanism to break down the precipitate. The ions remain in solution but do not affect the precipitate.

3. **Dilute H₂SO₄**: Dilute sulfuric acid also does not dissolve AgCl. The solubility product constant (Ksp) for AgCl indicates that it remains insoluble in dilute acids.

4. **Ammonium hydroxide (NH₄OH)**: AgCl is soluble in ammonium hydroxide due to the formation of a complex ion. When AgCl is treated with ammonium hydroxide, it forms a soluble complex ion:

\[ \text{AgCl} (s) + 2 \text{NH}_4\text{OH} (aq) \rightarrow [\text{Ag(NH}_3)_2]^+ (aq) + \text{Cl}^- (aq) + 2 \text{H}_2\text{O} (l) \]

### Step 4: Conclusion
From the analysis above, we conclude that the white precipitate of silver chloride (AgCl) is soluble in ammonium hydroxide.

### Final Answer
**Ammonium hydroxide**

---