Question

A cupboard of chemical opened and found four bottles containing water solutions, each of which had lost its label. Bottles 1, 2 and 3 contained colourless solutions, 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 and 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, the bottle 3 contains__________ .

• A. Copper (II) sulphate
• B. Hydrochloric acid
• C. Lead nitrate
• D. Sodium carbonate

Answer 1

The correct option is D Sodium carbonate

As bottle 2 and bottle 3 gives colourless and odourless gas, it may be carbon dioxide.

Generally, carbonates are decomposed by acids giving $C{O}_2$ gas. It suggests that bottle 2 and 3 contain sodium carbonate and $HCl$.

Bottle 3 and 4 gives blue precipitate which confirms the $C{u}^{2+}$ in either of bottles. $CuS{O}_4,CuC{l}_2,$ and $Cu(N{O}_3{)}_2$ are soluble and $CuC{O}_3$ is insoluble in water as evident from the reaction.

$C{u}^{2+}+C{O}_3^{2-}\to CuC{O}_3\downarrow \left(blueppt\right)$.

Thus, blue precipitate must be of copper carbonate.

Hence, bottle 4 is $CuS{O}_4$, 3 is $N{a}_{2}C{O}_3$, 2 is $HCl$ and 1 is $Pb(N{O}_3{)}_2$ as it gives white precipitate of $PbC{l}_2$ with bottle 2.

$PbN{O}_3,N{a}_{2}C{O}_3$ and $HCl$ solution are colourless but $CuS{O}_4$ is blue colour solution.

Therefore, option D is correct.