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A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless solution, while bottle 4 contained a blue solution.The labels from the bottles were lying scattered on the floor of the cupboard.They were: copper (II) sulphate, Hydrochloric acid lead nitrate , Sodium carbonate By mixing samples of the contents of the bottles, in pairs , the chemist made the following observations : Bottle 1 +Bottle 2 to White precipitate is formed. Bottle 1 +Bottle 3 to White precipitate is formed. Bottle 1 +Bottle 4 to White precipitate is formed. Bottle 2 +Bottle 3 to Colourless and odourless gas is evolved. Bottle 2 +Bottle 4 to No visible reaction is observed. Bottle 3 +Bottle 4 to Blue precipitate is formed. With the help of the above observations answer the following questions. Bottle 3 contains :

Answer»

COPPER (II) sulphate
Hydrochloric acid
Lead nitrate
Sodium carbonate

Solution :As bottle 2+ bottle 3 gives colourless and odourless gas, it may be carbon dioxide. Generally CARBONATES are decomposed by acids giving `CO_2` gas.It suggests that bottle 2 and 3 contains sodium carbonate and HCl. Bottle 3+4 gives blue precipitate which confirms the `Cu^(2+)` in either of bottles, `CuSO_4 , CuCl_2 and Cu(NO_3)_2`are soluble and `CuCO_3` is insoluble in water as evident from the reaction.
`Cu^(2+)+CO_3^(2-)toCuCO_3 DARR` (blue).Thus blue precipitate MUST be of copper carbonate.
Hence, bottle 4 is `CuSO_4` , 3 is `Na_2CO_3`, 2 is HCl (from above ) and 1 is `Pb(NO_3)_2` as it gives WHITE precipitate of `PbCl_2` with bottle (2).


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