An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution, a black precipitate is obtained. The substance is :

Options

A. `Hg^(2+)` salt
B. `Cr^(3+)` salt
C. `Ag^(+)` salt
D. `Pb^(2+)` salt

Explanation

To solve the problem step by step, we need to analyze the information given in the question regarding the substance and its reactions with hydrochloric acid and hydrogen sulfide.

### Step 1: Identify the Reaction with Dilute Hydrochloric Acid
When an aqueous solution of the substance is treated with dilute hydrochloric acid (HCl), it forms a white precipitate. This indicates that the substance is likely a metal salt that can form an insoluble chloride.

**Hint:** Think about common metal chlorides that are insoluble in water.

### Step 2: Determine the Nature of the Precipitate
The white precipitate formed when treated with dilute HCl is likely lead(II) chloride (PbCl2), which is known to be white and insoluble in cold water but soluble in hot water.

**Hint:** Recall the solubility rules for chlorides and identify which metal forms a white precipitate with HCl.

### Step 3: Analyze the Behavior of the Precipitate on Heating
The problem states that the white precipitate dissolves upon heating. This characteristic is specific to lead(II) chloride (PbCl2), which is soluble in hot water.

**Hint:** Consider the solubility of metal chlorides at different temperatures.

### Step 4: Reaction with Hydrogen Sulfide
When hydrogen sulfide (H2S) is passed through the hot solution, a black precipitate is obtained. This indicates the formation of lead(II) sulfide (PbS), which is known to be black.

**Hint:** Think about the typical reactions of metal ions with sulfide ions and the colors of the resulting precipitates.

### Step 5: Conclusion
Based on the reactions and the characteristics of the precipitates formed, the substance in question is lead(II) ions (Pb²⁺).

**Final Answer:** The substance is lead(II) chloride (PbCl2), which forms a white precipitate with dilute HCl and leads to the formation of a black precipitate of lead(II) sulfide (PbS) when treated with hydrogen sulfide.

### Summary of Steps:
1. Identify the formation of a white precipitate with dilute HCl.
2. Recognize that the white precipitate is likely PbCl2.
3. Note that PbCl2 dissolves upon heating.
4. Understand that passing H2S through the solution forms a black precipitate (PbS).
5. Conclude that the substance is lead(II) ions (Pb²⁺).