AS Chemistry 9701: Testing for Halide Ions with Silver Nitrate (Group 17)

A technician has three unlabelled bottles, each containing a colourless aqueous solution of a different sodium halide: sodium chloride, sodium bromide and sodium iodide. To identify the solution in bottle Q, the technician first acidifies a sample with dilute nitric acid, then adds aqueous silver nitrate. A yellow precipitate forms. Excess concentrated aqueous ammonia is then added to this precipitate, and it does not dissolve.

(a) Identify the halide ion present in bottle Q, giving your reasoning. [2]

(b) Write an ionic equation, including state symbols, for the formation of the precipitate in bottle Q. [2]

(c) Describe how the results of this same test (dilute nitric acid, then aqueous silver nitrate, then aqueous ammonia) would allow the technician to distinguish between solutions of sodium chloride and sodium bromide, referring to both the colour of each precipitate and its solubility in ammonia. [3]

(d) Explain why dilute nitric acid is added to each sample before the silver nitrate solution. [1]

Part (a): Identifying the halide in bottle Q

The three sodium halides give silver halide precipitates of different colours and different solubilities in aqueous ammonia:

Halide ion	Precipitate colour	Dissolves in dilute NH $_3$ (aq)?	Dissolves in concentrated NH $_3$ (aq)?
$\text{Cl}^-$	white	yes	yes
$\text{Br}^-$	cream	no	yes
$\text{I}^-$	yellow	no	no

Bottle Q gives a yellow precipitate that does not dissolve even in excess concentrated ammonia. This matches silver iodide exactly, so bottle Q contains sodium iodide (the iodide ion, $\text{I}^-$ ).

Part (b): Ionic equation for the precipitate

The precipitate forms from the direct combination of silver ions and iodide ions in solution:

$\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}$

Part (c): Distinguishing chloride from bromide

Applying the same acidified silver nitrate test, followed by aqueous ammonia:

Sodium chloride gives a white precipitate of $\text{AgCl}$ , which dissolves in dilute aqueous ammonia (forming the soluble complex ion $[\text{Ag(NH}_3)_2]^+$ ).
Sodium bromide gives a cream precipitate of $\text{AgBr}$ , which does not dissolve in dilute ammonia, but does dissolve in excess concentrated aqueous ammonia.

So the two solutions can be told apart both by the precipitate’s colour (white vs. cream) and by how much ammonia is needed to dissolve it (dilute vs. concentrated).

Part (d): Why dilute nitric acid is added first

Aqueous silver nitrate does not only react with halide ions. It would also form a precipitate with other anions that might be present in solution, such as carbonate ions ( $\text{Ag}_2\text{CO}_3$ ) or hydroxide ions ( $\text{AgOH}$ ), giving a false-positive result that could be mistaken for a halide precipitate. Adding dilute nitric acid first removes/reacts with these interfering ions (e.g. releasing $\text{CO}_2$ from carbonate), while the silver halide precipitates themselves are insoluble in dilute nitric acid, so they are unaffected and the test remains a reliable indicator of the halide present.

Final answers

(a) Bottle Q contains sodium iodide (iodide ion), identified by the yellow precipitate that is insoluble even in concentrated ammonia.
(b) $\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}$
(c) Chloride: white precipitate, dissolves in dilute ammonia. Bromide: cream precipitate, insoluble in dilute ammonia but dissolves in concentrated ammonia.
(d) Dilute nitric acid removes ions (e.g. carbonate, hydroxide) that would otherwise give a false-positive precipitate with silver ions, without dissolving the silver halide precipitates themselves.