On heating the salt with `NHO_(3) `and ammonium molybdate , formation of yellow ppt indicates the presence of (a)_____or (b) _____.

Answer

To solve the question regarding the formation of a yellow precipitate upon heating a salt with HNO3 and ammonium molybdate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The question states that we are heating a salt with HNO3 (nitric acid) and ammonium molybdate.

2. **Understand the Reaction**: When ammonium molybdate reacts with certain anions in the presence of nitric acid, it can form a yellow precipitate. The key anions of interest here are phosphate (PO4^3-) and arsenate (AsO4^3-).

3. **Write the Reaction for Phosphate**:
- When phosphate ions react with ammonium molybdate in an acidic medium, they form ammonium phosphomolybdate, which is a yellow precipitate.
- The balanced reaction can be represented as:
\[ H3PO4 + NH4^+ + MoO4^{2-} \rightarrow NH4^3PO4 \cdot 12MoO3 + H2O \]

4. **Write the Reaction for Arsenate**:
- Similarly, arsenate ions can also react with ammonium molybdate to form ammonium arsenomolybdate, which is also a yellow precipitate.
- The balanced reaction can be represented as:
\[ H3AsO4 + NH4^+ + MoO4^{2-} \rightarrow NH4^3AsO4 \cdot 12MoO3 + H2O \]

5. **Conclusion**: The formation of a yellow precipitate indicates the presence of either phosphate ions (PO4^3-) or arsenate ions (AsO4^3-).

### Final Answer:
(a) Phosphate (PO4^3-) or (b) Arsenate (AsO4^3-).

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