IGCSE Chemistry 0620: Distinguishing Chloride, Bromide and Iodide Ions

A water-treatment laboratory keeps three stock solutions, potassium chloride, potassium bromide and potassium iodide, for calibrating detection equipment. After a labelling error, three sample bottles are left marked only R, S and T, and a technician must use aqueous silver nitrate to work out which bottle contains which solution.

(a) Describe the test the technician should carry out, including how each sample should be prepared before aqueous silver nitrate is added. [2]

(b) The technician obtains the following results:

Bottle R: cream precipitate
Bottle S: white precipitate
Bottle T: yellow precipitate

Identify the solution in each of bottles R, S and T. [3]

(c) Explain why the technician adds dilute nitric acid to each sample before adding aqueous silver nitrate, rather than testing the samples directly with aqueous silver nitrate. [2]

Part (a): Carrying out the test

For each sample, the technician should:

Add dilute nitric acid to the sample.
Add aqueous silver nitrate.
Observe the colour of any precipitate formed.

Part (b): Identifying each bottle from the precipitate colour

Acidified aqueous silver nitrate gives a precipitate with each halide ion, and the colour identifies which halide is present:

Halide ion	Precipitate colour
chloride, $\text{Cl}^-$	white
bromide, $\text{Br}^-$	cream
iodide, $\text{I}^-$	yellow

$\text{Ag}^+(aq) + \text{X}^-(aq) \rightarrow \text{AgX}(s)$

Matching the results in the question to this table:

Bottle R (cream precipitate) $\rightarrow$ potassium bromide.
Bottle S (white precipitate) $\rightarrow$ potassium chloride.
Bottle T (yellow precipitate) $\rightarrow$ potassium iodide.

Part (c): Why the sample is acidified with nitric acid first

Aqueous silver nitrate does not only react with halide ions. It also forms precipitates with some other anions that might be present in a sample, such as carbonate ions:

$2\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{Ag}_2\text{CO}_3(s)$

If the sample were tested directly, without acidifying it first, a precipitate like this could form even in the absence of any halide ion, wrongly suggesting a halide is present. Adding dilute nitric acid first destroys carbonate ions (releasing carbon dioxide gas) and other similarly interfering ions, so that any precipitate later formed with silver nitrate can be confidently attributed to a halide ion rather than to some other anion in the sample.

Final answers

(a) Add dilute nitric acid, then aqueous silver nitrate; observe the precipitate colour.
(b) R: $\boxed{\text{potassium bromide}}$ ; S: $\boxed{\text{potassium chloride}}$ ; T: $\boxed{\text{potassium iodide}}$ .
(c) Nitric acid removes ions (such as carbonate ions) that would otherwise give their own precipitate with silver nitrate and be mistaken for a halide, so acidifying first prevents a false-positive result.