Question

A solution containing a mixture of metal cations was treated with dilute HCl and a precipitate formed. The solution was filtered, and H2S was bubbled through the acidic solution. A precipitate again formed and was filtered off. Then, the pH was raised to about 8 and H2S was again bubbled through the solution. This time, no precipitate formed. Finally, the solution was treated with a sodium carbonate solution, which resulted in the formation of a precipitate.
Classify the metal ions based on whether they were definitely present, definitely absent, or whether it was possible they were present in the original mixture.

Cd²⁺, Sb³⁺, Hg²⁺, Pb²⁺, Ba²⁺, Zn²⁺, Fe²⁺, Cr³⁺, Sn⁴⁺, Li⁺, Mg²⁺, Hg₂²⁺, Ag⁺

Answer 1

Answer:

Cd²⁺, Sb³⁺, Hg²⁺, Pb²⁺, Ba²⁺, Sn⁴⁺, Mg²⁺, Hg₂²⁺, Ag⁺- definitely present

Li⁺, Fe2+, Cr3+, Zn2+- definitely absent.

Explanation:

Ag⁺,Pb²⁺,Hg₂²⁺ could form precipitate with HCl in step one. Hence these ions are likely to be present

Cd²⁺, Sb³⁺, Sn⁴⁺, Hg²⁺ could form precipitate with H2S in step two. These ions are likely to be present.

Fe²⁺,Cr³⁺,Zn²⁺ these ions could form precipitates under basic conditions and insoluble sulphides in excess H2S but the question stated that no precipitate was formed in this step hence these cations are absent.

Mg²⁺,Ba²⁺ form precipitates with carbonate ion, they are likely to be present.