Match the Column — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Treatment of the aqueous solution with dilute HCl yields a white precipitate. Lead(II) chloride (PbCl2) is uniquely soluble in hot water but insoluble in cold water. Step 2: Upon heating, PbCl2 dissolves back into solution. When H2S is passed through this hot acidic solution containing Pb2+ ions, it precipitates as highly insoluble black lead(II) sulfide (PbS). Step 3: This characteristic behavior (HCl precipitation, hot water solubility, and H2S black precipitate formation) uniquely identifies the lead(II) ion (Pb2+), matching option (d).