A solution containing a mixture of metal cations was treated with dilute HCl and no precipitate formed. Next, H2S was bubbled through the acidic solution. A precipitate formed and was filtered off. Then, the pH was raised to about 8 and H2S was again bubbled through the solution. A precipitate again formed and was filtered off. Finally, the solution was treated with a sodium carbonate solution, which resulted in no precipitation. Which metal ions were definately present, which were definitely absent, and which may or may not have been present in the original mixture?Ag+, Ba2+, Mg2+, Hg2+, Pb2+, Hg2^2+, Cu2+, Zn2+, Ni2+, Co2+, Sn2+, Li+, Sb3+

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Xema8 Xema8 · Answer 1 · 2019-08-22T13:34:15+02:00

Answer:

Explanation:

Ions which will be definitely absent -

Ag⁺ , Hg₂⁺ , Pb⁺², because their chlorides are insoluble in water.

Ba⁺² ,Ca⁺² because their carbonates are insoluble in water,

Ions which will be definitely present

Cu⁺² , Pb⁺² . Hg⁺² , Sb⁺³ , Sn⁺² , because their sulphides are insoluble at pH = 8.

For the rest nothing can be said.

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