A solution of AgCl in NH4OH gives a white precipitate when acidified with HNO3. — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: In the first reaction, Cu2+ oxidizes I- to triiodide ions (I3-) in the presence of excess iodide, while being reduced to Cu+ which precipitates as copper(I) iodide: 2Cu2+ + 5I- -> 2CuI↓ + I3-. Thus, [X] is I3-. Step 2: Triiodide (or iodine in KI) is titrated with sodium thiosulfate (S2O32-), which reduces it back to iodide (I-): I3- + 2S2O32- -> 3I- + S4O62-. Thus, [Y] is I-. Step 3: This identifies X as I3- and Y as I-, matching option (a).