When copper sulphate solution is treated with potassium iodide and excess of hypo solution is added — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Cu2+ reacts with iodide ions (I-) to undergo a redox reaction: 2Cu2+ + 4I- -> 2CuI↓ (white precipitate) + I2 (brown solution). Step 2: Upon adding sodium thiosulfate (hypo solution, Na2S2O3), the brown triiodide/iodine is reduced to colorless iodide (I-), making the white precipitate of copper(I) iodide (CuI) clearly visible. Step 3: Therefore, the white precipitate remaining in the solution is CuI, corresponding to option (c).