Question

A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially soluble in hot water. The supernatant and the precipitate are placed into two separate test tubes. The solid remaining after treatment with hot water turns black on addition of 6 M NH3. The supernatant is tested with K2CrO4 and a yellow precipitate forms. No other precipitates are observed while performing the procedure. Which of the ions are present and which are absent? State your reasoning with equations

Answer 1

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$.

• Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
• Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$.
• Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$. The latter precipitate is yellow.