When phosphate radical with ammonium molybdate, the colour of precipitate obtained as

Options

A. Green
B. Pink
C. Canary yellow
D. Violet

Answer

To determine the color of the precipitate formed when the phosphate radical reacts with ammonium molybdate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are the phosphate radical (PO₄³⁻) and ammonium molybdate (NH₄)₂[MoO₄].

2. **Write the Reaction**:
The reaction occurs in the presence of nitric acid (HNO₃). The balanced chemical equation for the reaction can be written as:
\[ \text{PO}_4^{3-} + \text{(NH}_4)_2\text{MoO}_4 \xrightarrow{\text{HNO}_3} \text{NH}_4\text{PO}_4\cdot12\text{MoO}_3 + \text{NH}_4\text{NO}_3 + \text{H}_2\text{O} \]

3. **Formation of Precipitate**:
The product formed in this reaction is ammonium phosphomolybdate, which is represented as:
\[ \text{NH}_4\text{PO}_4\cdot12\text{MoO}_3 \]

4. **Identify the Color of the Precipitate**:
The precipitate formed, ammonium phosphomolybdate, is known to have a characteristic color. It is typically a **canary yellow** precipitate.

5. **Conclusion**:
Therefore, the color of the precipitate obtained when the phosphate radical reacts with ammonium molybdate is **canary yellow**.