A white ppt obtained in the anylsis of a mixture becomes black on treatment with `NH_(4)OH` it may be

Options

A. `Hg_(2)CI_(2)`
B. `HgCI_(2)`
C. `PbCI_(2)`
D. `AgCI`

Answer

To solve the problem of identifying the white precipitate that turns black upon treatment with ammonium hydroxide (NH₄OH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the White Precipitate**:
The question states that a white precipitate is formed. A common white precipitate in inorganic salt analysis is mercurous chloride (Hg₂Cl₂), also known as calomel.

2. **Treatment with Ammonium Hydroxide**:
When mercurous chloride (Hg₂Cl₂) is treated with ammonium hydroxide (NH₄OH), a chemical reaction occurs.

3. **Chemical Reaction**:
The reaction can be represented as:
\[ \text{Hg}_2\text{Cl}_2 + 2 \text{NH}_4\text{OH} \rightarrow \text{HgO} \cdot \text{HgNH}_2\text{Cl} + 2 \text{NH}_4\text{Cl} + \text{H}_2\text{O} \]
In this reaction, the white precipitate (Hg₂Cl₂) reacts with ammonium hydroxide to form a black precipitate, which is a complex of mercury.

4. **Formation of Black Precipitate**:
The black precipitate formed is known as Amido Mercuric Chloride (HgO·HgNH₂Cl). This is the compound that gives the black color upon the addition of NH₄OH.

5. **Conclusion**:
Therefore, the white precipitate that becomes black upon treatment with ammonium hydroxide is mercurous chloride (Hg₂Cl₂), and the black precipitate formed is Amido Mercuric Chloride.

### Final Answer:
The white precipitate is Hg₂Cl₂ (Mercurous Chloride), which becomes black (Amido Mercuric Chloride) upon treatment with NH₄OH.

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