Chemistry, asked by lihtaiwangsu, 3 months ago

how will you distinguish between following pairs of compounds using NH4OH.i)copper sulphate and iron.ii)sulphate.ii)zinc nitrate and lead nitrate.iii)iron sulphate and iron iii sulphate

Answers

Answered by farhaanaarif84
2

Answer:

Add NaOH solution in excess to the two solutions. The one in which white ppt. initially formed dissolves in excess of NaOH solution is Zn(NO3)2 solution and the other is Ca(NO3)2 solution.

2. Add freshly prepared ferrous sulphate solution to the two solutions. Then by the side of the test tube, pour cone, sulphuric acid to each slowly. The one in which brown ring appears is sodium nitrate solution while the other is sodium chloride solution.

3. Add NaOH solution to both the solutions. The one which gives a reddish brown ppt. is iron(III) chloride solution and the one which gives blue ppt. is copper chloride solution.

4. When sodium hydroxide solution is added to iron (II) sulphate solution, a dirty green precipitate is formed.

When sodium hydroxide solution is added to copper (II) sulphate solution, light blue precipitate is formed.

5. When ammonium hydroxide is added to zinc nitrate solution, a white precipitate is formed. The white precipitate dissolves when excess of ammonium hydroxide is added.

When ammonium hydroxide is added to calcium nitrate solution, no visible reaction occurs even with the addition of excess of NH4OH.

6. When sodium hydroxide is added to Iron (II) chloride, dirty green precipitate of Fe(OH)2 is formed.

FeCl2 + 2 NaOH ⟶ Fe (OH)2 ↓ + 2NaCl

When sodium hydroxide is added to iron (III) chloride, reddish brown precipitate is formed

FeCl3 + 2NaOH ⟶ Fe(OH)3 + 3NaCl

Answered by aryanthakur00000000
1

Answer:

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