Testing for chlorides (Cl^-), bromides (Br^-) and iodides (I^-)
4 marks · NormalDescribe the steps you would take to test a sample for iodide ions using silver nitrate. Include all reagents added and any important precautions.
What result would indicate that iodide ions are present?
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) — · must show Dissolve sample in water (if solid), Add dilute nitric acid, Add silver nitrate solution, Mix and observe for a precipitate, Order/importance of adding nitric before silver nitrate
(b) Yellow precipitate forms · must show yellow precipitate, silver iodide
Must show: steps: make solution, add nitric acid, add silver nitrate, safety/precaution about order
Also accept: Acidify with nitric acid before silver nitrate; allow mention of dissolving the sample and observation after mixing.
Common mistakes: Omitting nitric acid; Adding hydrochloric acid instead of nitric acid; Stating only the observation without any method
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Marking points
M1 (a) · 1 mark · Dissolve sample in water (if a solid) OR use a solution sample.
M1 (a) · 1 mark · Add dilute nitric acid to the sample (acidification), explaining or implying removal of interfering ions.
M1 (a) · 1 mark · Add silver nitrate solution after acidification (order is important).
B1 (b) · 1 mark · Yellow precipitate observed (or stated as silver iodide forms).
Worked solution
a) Dissolve a small amount of the sample in distilled water if it is not already in solution.
Add a few drops of dilute nitric acid to the solution to acidify it. This removes any interfering ions, such as carbonates, that could give a false result.
Add a few drops of silver nitrate solution to the acidified sample.
b) If iodide ions are present, a yellow precipitate of silver iodide will form.