Question

When 6M HCl is added to the solution, a white precipitate forms. What cation(s) may be present in the solution?

Answer 1

The white precipitate formed upon adding 6M HCl suggests the presence of cations like silver, lead, or mercury(I), which react with chloride ions to form insoluble compounds such as AgCl, PbCl2, or Hg2Cl2, respectively. Barium could also be a possibility if sulfate is the anion.

Step-by-step explanation:

When 6M HCl is added to a solution and a white precipitate forms, it suggests the presence of cations that form insoluble chlorides. A common cation that forms a white precipitate with chloride ions is silver (Ag+). Similarly, lead (Pb2+) and mercury(I) (Hg22+) can also form white precipitates with chloride ions. For example, when Hg2Cl2 is treated with ammonia, it forms Hg and Hg(NH2)Cl. If excess HCl is added to a solution containing silver ions, a white precipitate of AgCl is formed, which darken upon exposure to light.

If the solution contained barium, sulfate could be a possible anion since barium sulfate forms a white precipitate that is insoluble in hydrochloric acid. Adding barium chloride (BaCl2) to a sulfate solution would result in the formation of a white precipitate of barium sulfate (BaSO4).

Answer 2

`Ag^+, Pb^(2+), Hg_2^(2+)`

Step-by-step explanation:

The majority of metal chloride salts are soluble, however, there are notable exceptions. When chloride ions are added to a solution containing lead(II) cations, silver cations or mercurous cations, precipitates are formed.

When silver cations react with chloride, we obtain an insoluble silver chloride:

`Ag^+ + Cl^-\rightarrow AgCl`

When lead(II) cations react with chloride anions, lead(II) precipitate is formed, however, it becomes soluble at higher temperatures:

`Pb^(2+) + 2 Cl^-\rightarrow PbCl_2`

Finally, the final precipitate would be:

`Hg_2^(2+) + 2 Cl^-\rightarrow Hg_2Cl_2`.