Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. NaCl solution and one fo

Question

Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. NaCl solution and one formed a black precipitate with H2S. The salt could be:

Prashant Kotian · Accepted Answer

Correct answer: B. Step 1: Precipitation with NaCl (Cl-) indicates the cation forms an insoluble chloride, which restricts the choices to Pb2+, Ag+, or Hg22+. Step 2: Reaction with excess ammonia yields a stable white hydroxide/basic salt precipitate with Pb2+, while Ag+ forms a soluble colorless complex [Ag(NH3)2]+ (no precipitate) and Hg22+/Hg2+ undergo blackening. Step 3: Reaction with H2S yields black lead(II) sulfide (PbS↓). Hence, the salt is lead(II) nitrate, Pb(NO3)2, matching option (b).

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