Question

A certain compound X when treated with copper sulphate solution yields a brown precipitate. On adding hypo solution, the precipitate turns white. The compoundXis

• A. $K_2CO_3$
• B. KI
• C. KBr
• D. $K_3PO_4$

Answer 1

The Correct Option is B

To determine which compound X, when treated with copper sulfate solution, yields a brown precipitate that turns white on adding hypo solution, we need to analyze the chemical reactions involved.

1. Firstly, when potassium iodide (KI) is added to a copper sulfate (CuSO₄) solution, the following reaction occurs:

2\text{KI} + \text{CuSO}_4 \rightarrow \text{CuI}_2 + \text{K}_2\text{SO}_4

2. The copper iodide (CuI₂) formed is unstable and decomposes to form CuI (a brown precipitate) and free iodine (I₂):

\text{CuI}_2 \rightarrow \text{CuI} + \frac{1}{2}\text{I}_2

3. Secondly, upon adding "hypo" solution (sodium thiosulfate, Na₂S₂O₃), the iodine reacts with it to form sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI), turning the brown precipitate white:

\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}

This reaction removes iodine, and the remaining precipitate is white (CuI).

4. Thus, the compound X is KI, as only potassium iodide will undergo the above reaction sequence, resulting in a brown precipitate that turns white upon addition of hypo solution.

Let's evaluate why other options are incorrect:

• K_2CO_3: Potassium carbonate does not react with copper sulfate to give a similar brown precipitate that turns white.
• KBr: While bromides can react with copper sulfate, they do not yield the same color change as iodides when treated with hypo solution.
• K_3PO_4: Potassium phosphate will not react with copper sulfate to give a precipitate that changes color as described.

Therefore, the correct answer is KI.