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Question

(a) State the colour of the residue formed when nitrates of
1. Calcium 2. Zinc 3. Lead 4. Copper are strongly heated.

(b) Give one test each to distinguish between the following pairs of chemicals.
1. Zinc nitrate solution and calcium nitrate solution.
2. Sodium nitrate solution and sodium chloride solution.
3. Iron (III) chloride solution and copper chloride solution.

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Solution

(a) The colour of the residues obtained after strongly heating nitrates of the following elements are given below:
1. Calcium - White
2. Zinc - Yellow when hot, white when cold
3. Lead - Lime yellow
4. Copper - Black

(b) Tests to distinguish between the following pairs of chemicals are as follows:

1. Zinc nitrate solution and calcium nitrate solution - Sodium hydroxide test
With sodium hydroxide solution, calcium nitrate will form an insoluble white precipitate. However, zinc nitrate will form a white precipitate which will be soluble in excess of sodium hydroxide.
Ca(NO3)2 + 2NaOH → Ca(OH)2 + 2NaNO3
Zn(NO3)2 + 2NaOH → Zn(OH)2 + 2NaNO3
Zn(OH)2 + 2NaOH → Na2[Zn(OH)4]

2. Sodium nitrate solution and sodium chloride solution - Addition of silver nitrate in presence of dilute nitric acid
When silver nitrate solution in presence of dilute nitric acid is added to the solution of sodium chloride, a white precipitate of silver chloride is obtained. This precipitate is soluble in ammonium hydroxide solution. However, no such reaction occurs with sodium nitrate.
NaNO3 + AgNO3 → No reaction
NaCl + AgNO3 → AgCl + NaNO3
AgCl + 2NH4OH → [Ag(NH3)2]Cl + 2H2O

3. Iron (III) chloride solution and copper chloride solution - Ammonium hydroxide test
Addition of ammonium hydroxide solution to iron (III) chloride solution gives a reddish brown precipitate of iron (III) hydroxide. However, with copper chloride solution a pale blue precipitate of copper hydroxide is formed. The complex dissolves in excess ammonium hydroxide and forms a deep blue coloured solution containing tetraamine copper complex.
FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl
CuCl2 + 2NH4OH → Cu(OH)2 + 2NH4Cl
Cu(OH)2 + 4NH4OH →[Cu(NH3)4](OH)2 + 4H2O

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