Sodium salt solution `+AgNO_(3)"soln."to`Coloured precipitate.
If coloured precipitate is soluble in both dil. `HNO_(3)` and excess conc. `NH_(3)` solution then which of the following anion is present in the salt solution?

Options

A. `S^(2-)(aq.)`
B. `I^(-)(aq.)`
C. `PO_(4)^(3-)(aq.)`
D. `Br^(-)(aq.)`

Answer

To determine which anion is present in the sodium salt solution based on the given conditions, we can follow these steps:

### Step 1: Identify the Reaction
When a sodium salt solution is mixed with AgNO3 (silver nitrate), a colored precipitate is formed. This indicates that a reaction is occurring between the anion in the sodium salt and AgNO3.

### Step 2: Analyze Possible Anions
We need to consider several common anions and their reactions with AgNO3:
- **Sulfide (S²⁻)**: Forms Ag2S (black precipitate)
- **Iodide (I⁻)**: Forms AgI (yellow precipitate)
- **Phosphate (PO4³⁻)**: Forms Ag3PO4 (yellow precipitate)
- **Bromide (Br⁻)**: Forms AgBr (pale yellow precipitate)

### Step 3: Solubility of Precipitates
The problem states that the colored precipitate is soluble in both dilute HNO3 and excess concentrated NH3. We need to check the solubility of the precipitates formed from the anions mentioned:
- **Ag2S** (from S²⁻): Insoluble in both HNO3 and NH3.
- **AgI** (from I⁻): Insoluble in both HNO3 and NH3.
- **Ag3PO4** (from PO4³⁻): Soluble in both HNO3 and NH3.
- **AgBr** (from Br⁻): Insoluble in HNO3 but soluble in NH3.

### Step 4: Conclusion
Since the only precipitate that is soluble in both dilute HNO3 and excess concentrated NH3 is Ag3PO4, we can conclude that the anion present in the sodium salt solution is **phosphate (PO4³⁻)**.

### Final Answer
The anion present in the sodium salt solution is **PO4³⁻ (phosphate ion)**.

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