(i) A white solid mixture of two salts containing a common cations in insoluble in water. It dissolves in dilute HCl producing some gases (with effervescence) that turn an acidified dichromate solution gren. After the gases are passed through the acidified dichromate solution, the emerging gas turns baryta water milky.
(ii) On treatment with dilute `HNO_(3)`, the white solid gives a solution which does not directly give a precipitate with a `BaCl_(2)` solution but gives a white precipitate when warmed with `H_(2)O_(2)` and then treated with a `BaCl_(2)` solution.
(iii) The solution of the mixture in dilute HCl, when treated with `NH_(4)Cl,NH_(4)OH and ` an `Na_(2)HPO_(4)` solution, gives a white precipitate.
Q. The white precipitate obtained in (ii) indicates the presence of a:

Options

A. carbonate
B. sulphide
C. sulphite
D. chloride

Answer

To solve the problem step by step, we need to analyze the information provided in the question and the video transcript.

### Step 1: Analyze the White Solid Mixture
The white solid mixture contains two salts with a common cation that is insoluble in water. When this mixture is treated with dilute HCl, it produces gases with effervescence. The gases produced turn an acidified dichromate solution green, indicating the presence of a reducing agent, likely sulfide ions (S²⁻) or carbonate ions (CO₃²⁻).

**Hint:** Look for gases that can reduce dichromate ions, such as H₂S or CO₂.

### Step 2: Identify the Gas Produced
After passing the gases through the acidified dichromate solution, the emerging gas turns baryta water milky. This indicates the presence of carbon dioxide (CO₂), which reacts with barium hydroxide (from baryta water) to form barium carbonate (BaCO₃), a white precipitate.

**Hint:** Consider the reactions that produce CO₂ and how it interacts with baryta water.

### Step 3: Treatment with Dilute HNO₃
When the white solid is treated with dilute HNO₃, it does not produce a precipitate with BaCl₂ directly, but does when warmed with H₂O₂ and then treated with BaCl₂. This suggests that the solid contains sulfite ions (SO₃²⁻) because sulfites do not precipitate with barium ions at room temperature but can oxidize to sulfate ions (SO₄²⁻) when treated with H₂O₂.

**Hint:** Identify the role of H₂O₂ in oxidizing sulfites to sulfates.

### Step 4: Formation of White Precipitate
When the solution is treated with BaCl₂ after oxidation, it forms a white precipitate of barium sulfate (BaSO₄). This confirms the presence of sulfite ions in the original mixture.

**Hint:** Remember that barium sulfate is a well-known white precipitate that confirms the presence of sulfate ions.

### Step 5: Final Analysis with NH₄Cl, NH₄OH, and Na₂HPO₄
The solution of the mixture in dilute HCl, when treated with NH₄Cl, NH₄OH, and Na₂HPO₄, gives a white precipitate. This indicates the presence of phosphate ions (PO₄³⁻) in the mixture.

**Hint:** Consider the reactions that lead to the formation of precipitates with ammonium salts and phosphates.

### Conclusion
The white precipitate obtained in step 2 indicates the presence of sulfite ions (SO₃²⁻) in the original mixture.

### Final Answer
The white precipitate obtained in (ii) indicates the presence of a sulfite ion (SO₃²⁻).