Question

A compound (X) when treated with \( \text{CuSO}_4 \) solution yields a brown precipitate. On adding hypo solution the precipitate turns white. The compound (X) is

• A. \( \text{KBr} \)
• B. \( \text{K}_2\text{CrO}_3 \)
• C. \( \text{KI} \)
• D. \( \text{K}_3\text{PO}_4 \)

Hint:

This is a standard volumetric/qualitative test for copper. Remember that \( \text{Cu}^{2+} \) does not form a stable \( \text{CuI}_2 \) because \( \text{I}^- \) is a strong reducing agent and easily reduces \( \text{Cu}^{2+} \) to \( \text{Cu}^+ \) while being oxidized to \( \text{I}_2 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem describes a sequence of redox reactions used in qualitative analysis and iodometric titrations.
Copper(II) ions ($Cu^{2+}$) are known to be oxidizing agents. Iodide ions ($I^-$), found in KI, are strong reducing agents.
The characteristic "brown precipitate" that turns "white" upon addition of "hypo" (sodium thiosulfate) is a definitive test for the presence of iodide and the formation of copper(I) species.
Step 2: Key Formula or Approach:
The reactions occur in two distinct stages:
1. Reaction with Copper Sulfate:
\[ 2Cu^{2+} + 4I^- \rightarrow 2CuI \downarrow (\text{white ppt}) + I_2 (\text{brown solution}) \]
2. Reaction with Hypo (Sodium Thiosulfate):
\[ I_2 + 2S_2O_3^{2-} \rightarrow 2I^- (\text{colorless}) + S_4O_6^{2-} \]
Step 3: Detailed Explanation:
When an aqueous solution of potassium iodide (KI) is added to copper(II) sulfate ($CuSO_4$), a redox reaction takes place immediately.
Copper(II) ions oxidize the iodide ions to elemental iodine ($I_2$). Simultaneously, the copper(II) ions are reduced to copper(I) ions.
The copper(I) ions react with excess iodide to form copper(I) iodide ($CuI$), which is an insoluble white solid.
However, the liberated iodine ($I_2$) is brown and dissolves in the remaining KI solution to form triiodide ions ($I_3^-$).
Visually, the white $CuI$ precipitate is masked by the intense brown color of the iodine solution, making the mixture appear as a "brown precipitate" or a dark brown suspension.
When sodium thiosulfate ($Na_2S_2O_3$, commonly called "hypo") is added, it acts as a reducing agent for the iodine.
Hypo reduces the brown $I_2$ back to colorless $I^-$ ions and is itself oxidized to the tetrathionate ion ($S_4O_6^{2-}$).
Once the iodine is completely reduced, its brown color disappears, and the white color of the $CuI$ precipitate becomes clearly visible. This explains the observation that the precipitate "turns white."
Other halides like $KBr$ do not reduce $Cu^{2+}$ to $Cu^+$ under standard conditions, so no precipitate or color change is observed with them.
Step 4: Final Answer:
The compound X is potassium iodide (KI), as it explains all the experimental observations. The correct option is (C).