An aqueous solution of a white salt A gives a white precipitate B on treatment with dilute HCl in cold condition. B is soluble in boiling water. An aqeous solution of A gives a yellow precipitate on treatment with a solution of `K_(2)CrO_(4)`. The soda extract of A is acidified with diolute `H_(2)SO_(4)`, boiled to remove `CO_(2)` and treated with a freshly prepared solution of `FeSO_(4)`. concentratd `H_(2)SO_(4)` is added to the resulting solution. a brownn ring is formed at the junction of the two layers.
Q. On treatment with a KI solution, an aqueous solution of A will give:

Options

A. a yellow precipitate, soluble in boiling water
B. a yellow precipitate, insoluble in boiling water
C. a white precipitate, soluble, soluble in boiling water
D. a white precipitate, insoluble in boiling water

Answer

To solve the problem step by step, we will analyze the information provided about the white salt A and its reactions.

### Step 1: Identify the White Salt A
The problem states that an aqueous solution of a white salt A gives a white precipitate B when treated with dilute HCl in cold conditions. The white precipitate B is soluble in boiling water.

**Hint:** Consider common white salts that form precipitates with dilute HCl.

**Solution:** The white salt A is likely to be lead(II) nitrate (Pb(NO3)2). When lead(II) nitrate reacts with dilute HCl, it forms lead(II) chloride (PbCl2), which is a white precipitate.

### Step 2: Confirm the Properties of Precipitate B
The precipitate B formed is soluble in boiling water.

**Hint:** Look for the solubility characteristics of common lead salts.

**Solution:** Lead(II) chloride (PbCl2) is indeed a white precipitate that is soluble in hot water, confirming that our identification of salt A as lead(II) nitrate is correct.

### Step 3: Analyze the Reaction with K2CrO4
The problem states that an aqueous solution of A gives a yellow precipitate when treated with a solution of K2CrO4.

**Hint:** Think about the possible reactions of lead salts with chromate ions.

**Solution:** When lead(II) nitrate reacts with potassium chromate (K2CrO4), it forms lead(II) chromate (PbCrO4), which is a yellow precipitate. This further supports that salt A is lead(II) nitrate.

### Step 4: Soda Extract and the Brown Ring Test
The soda extract of A is acidified with dilute H2SO4, boiled to remove CO2, and treated with freshly prepared FeSO4. The addition of concentrated H2SO4 results in a brown ring at the junction of the two layers.

**Hint:** Recall the significance of the brown ring test in inorganic chemistry.

**Solution:** The formation of a brown ring indicates the presence of nitrate ions (NO3-) in the solution, confirming that lead(II) nitrate is indeed the salt A.

### Step 5: Reaction with Potassium Iodide (KI)
Now, we need to determine what happens when an aqueous solution of A (lead(II) nitrate) is treated with potassium iodide (KI).

**Hint:** Consider the reactivity of lead(II) ions with iodide ions.

**Solution:** When lead(II) nitrate reacts with potassium iodide, it forms lead(II) iodide (PbI2), which is a yellow precipitate, along with potassium nitrate (KNO3). The reaction can be summarized as:
\[ \text{Pb(NO}_3\text{)}_2 + 2 \text{KI} \rightarrow \text{PbI}_2 \downarrow + 2 \text{KNO}_3 \]

### Final Answer
On treatment with a KI solution, an aqueous solution of A will give a yellow precipitate of lead(II) iodide (PbI2).

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