Exp 6.1 A - To detect the presence of chloride ions in aqueous solution

Question 1

Apparatus required (3)

Answer 1

test tubes, dropper, spatula, wash bottle

Question 2

chemicals required (4)

Answer 2

any soluble chloride salt (e.g. KCl, NaCl, etc..) silver nitrate solution, dilute ammonia solution, deionised water

Question 3

Explain all 4 steps of the procedure

Answer 3

1. Fill a test tube to 1/4 full with deionised water in a wash bottle.
2. Add a small amount of the chloride salt (NaCl) to the water. Shake to help the salt dissolve. A clear solution should be obtained.
3. Add a few drops of silver nitrate solution (AgNO3) using a dropper. The clear solution turns cloudy as a white precipitate is formed.
4. Add 1/4 test tube of dilute ammonia solution and shake gently. The cloudiness disappears.

Question 4

What is observed when a few drops of silver nitrate solution (AgNO3) is added to the chloride salt?
Explain why this happens.

Answer 4

The solution turns cloudy and a white precipitate is formed.
This is silver chloride.
It is formed when the silver ions from the silver nitrate combine with the chloride ions from the chloride salt that are dissolved in the water.
Ag+ + Cl- —> AgCl↓

Question 5

Why is cloudiness observed?

Answer 5

because the silver chloride formed doesn’t dissolve in water

Question 6

What does this symbol indicate? ↓

Answer 6

The arrow pointing down as shown indicates a precipitate is formed

Question 7

What is the conclusion of this experiment?

Answer 7

Chloride ions were present in solution

Question 8

What are the reagents in this experiment?

Answer 8

the silver nitrate