On adding KI solution in excess to a solution of CuSO4 we get a precipitate 'P' and another liquor ' — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Cu2+ ions react with iodide (I-) to form unstable copper(II) iodide (CuI2), which spontaneously decomposes to precipitate white copper(I) iodide (CuI, precipitate 'P') and liberate free iodine (I2). Step 2: In the presence of excess KI, the liberated iodine (I2) dissolves by reacting with iodide to form soluble triiodide ions (KI3) in the liquor ('M'). Step 3: Thus, precipitate 'P' is CuI and liquor 'M' is KI3 solution, matching option (c).