Question

To your unknown you add HCl to precipitate the Group A ions (Ag and Hg). This precipitate is removed and NH3 is added to it to redissolve the Ag and check for the presence of Hg. From the picture below which was taken right after the NH3 was added, please select a choice that best represents what you are seeing.

Answer 1

Options are not given, however, the following reactions occurs when Group A ions are reacted with HCl followed by NH3

Answer:

Gray precipitate is seen, which confirms the presence of mercury ions

Explanation:

Selective precipitation is a qualitative analysis, which involves addition of a carefully selected reagents to an aqueous mixture of ions. This results in the precipitation of one or more ions, while leaving the rest in solution. Later, a reaction specific to that ion is carried out separately to determine its identity.

HCl react with both Ag+ and Hg+ ions to form the following precipitates:

Ag+(aq) + Cl-(aq) → AgCl(s)

$Hg_{2}^{2+}$(aq) + 2Cl- → $Hg_{2}Cl_{2}$(s)

The precipitate, i.e silver chloride and mercury(I) chloride is removed and solution of NH3 is added.

Silver chloride will dissolve since its forms a soluble complex ion:

 AgCl(s) + $2NH_{3}$(aq) → $Ag(NH_{3})_{2}^{+}(aq) + Cl^{-}(aq)$

However, Mercury(I) chloride will react with ammonia to form a gray solid which is actually a mixture of black mercury and white $AgNH_{2}Cl$ :

$Hg_{2}Cl_{2}$(s) + $2NH_{3}$(aq) → Hg(l) + $HgNH_{2}Cl(s) + NH_{4}^{+}(aq) + Cl^{-}(aq)$

The presence of gray solid is the confirmation of the presence of $Hg_{2}^{2+}$ ion