Question

1 after the precipitation of the group 1 cations as chlorides by the reaction Hcl nearly all of the precipitate dissolves in the hot water which ion or ions are present and which are absent

2a what type of reaction causes the formation of metallic mercury?
2b what reagent is responsible for this reaction?

3a what reagent causes Dooley chloride to dissolve into solution? give the formulas of the silver complex that is formed
3b what reagent causes the silver chloride precipitate to reform from the aqueous silver complex? why does this Happen?

i need help with this homework its due in forty five mins please help.

Answer 1

1) Group One analysis -- Ag+, Pb2+, and Hg2^2+

=> In the first step you precipitated all as chlorides: AgCl(s), PbCl2(s), Hg2Cl2(s).

=>The PbCl2 is separated based on its solubility in hot water. That leaves a mixture of AgCl(s) and Hg2Cl2(s).

2a) Disproportionation reaction causes the formation of metallic mercury.

2b) To the above mixture ammonia is added. The ammonia causes the disproportionation of Hg2Cl2 to form black Hg metal and white HgNH2Cl, giving a gray precipitate.

Hg2Cl2(s) + 2NH3(aq) --> Hg(l) + HgNH2Cl(s) + NH4Cl(aq)

3) Aqueous ammonia also causes AgCl to dissolve by forming a complex ion.

AgCl(s) + 2NH3(aq) ==> Ag(NH3)2^+(aq) + Cl^-(aq)

So ammonia causes the silver chloride precipitate to reform from the aqueous silver complex.