AQA A-Level Inorganic Chemistry 3.2.3 Group 7 Tests for Halide Ions
3.2.3

Tests for Halide Ions

Learn how to test for chloride, bromide, and iodide ions using silver nitrate and ammonia, and explore the use of chlorine in water treatment.

Testing for the presence of halide ions in aqueous solutions is a fundamental practical technique in chemistry. Because halide salts look identical as colourless solutions, chemical tests are used to trigger distinct physical changes that allow us to identify chloride, bromide, and iodide ions. Additionally, this topic covers the industrial chemistry of chlorine, focusing on its role in water chlorination.

🔑 Key Principle: Testing Sequence

To identify a halide ion, you first acidify the sample with dilute nitric acid, then add silver nitrate solution to form a silver halide precipitate. If the colours of the precipitates are too similar to distinguish, you carry out confirmatory solubility tests using dilute and concentrated ammonia solution.

1. The Silver Nitrate Test

To test for halide ions in solution, you must follow a two-step procedure:

  1. Acidification: Add a few drops of dilute nitric acid (\(\text{HNO}_3\)). This is crucial because it removes any interfering carbonate (\(\text{CO}_3^{2-}\)) or sulfate (\(\text{SO}_4^{2-}\)) ions that would otherwise react with silver ions to produce a false positive white precipitate.
  2. Precipitation: Add a few drops of silver nitrate solution (\(\text{AgNO}_3\)). The silver ions react with the halide ions to form insoluble silver halide precipitates:
\[ \text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX}(\text{s}) \]

The colours of the precipitates are:

The ionic equations for these precipitation reactions are:

\[ \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s}) \]

\[ \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s}) \]

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

📝 AQA Examiner Tip

You must never acidify the solution with hydrochloric acid (\(\text{HCl}\)) or sulfuric acid (\(\text{H}_2\text{SO}_4\)). Hydrochloric acid contains chloride ions, which would form a white precipitate of \(\text{AgCl}\) and ruin the test. Sulfuric acid contains sulfate ions, which would react with silver ions to form a white precipitate of silver sulfate, causing a false positive. Only use dilute nitric acid.

2. Confirmatory Ammonia Solubility Tests

In practice, white, cream, and yellow precipitates can look very similar, especially under laboratory lighting. To confirm the identity of the halide, we test the solubility of the silver halide precipitates in aqueous ammonia (\(\text{NH}_3\)):

📖 Definition: Precipitate

An insoluble solid that emerges from a liquid solution during a chemical reaction.

This trend in solubility down the group relates to the increasing size of the halide ion. As the halide ion becomes larger, the ionic bond in the silver halide lattice becomes stronger in terms of covalent character (silver iodide has the most covalent character), making it harder for the ammonia ligand to break the lattice and form a soluble complex ion, \([\text{Ag}(\text{NH}_3)_2]^+\).

Testing for Halide Ions with Silver Nitrate and Ammonia Chloride (Cl⁻) Bromide (Br⁻) Iodide (I⁻) STEP 1: ADD ACIDIFIED AgNO₃(aq) AgCl White ppt AgBr Cream ppt AgI Yellow ppt STEP 2: ADD AQUEOUS AMMONIA (NH₃) Dissolves in dilute NH₃ Dissolves in conc. NH₃ only Insoluble in conc. NH₃

3. Chlorine in Water Treatment

Chlorine is added to drinking water and swimming pools to kill bacteria, viruses, and other pathogens that cause waterborne diseases like cholera. When chlorine is added to water, it undergoes a reversible reaction:

\[ \text{Cl}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{HCl}(\text{aq}) + \text{HClO}(\text{aq}) \]

In this reaction, chlorine undergoes disproportionation. This is a type of redox reaction where a single element is simultaneously oxidised and reduced.

📖 Definition: Disproportionation

A redox reaction in which an element in a single chemical species is simultaneously oxidised and reduced to form two different products with different oxidation states.

Let's verify this by checking the oxidation states of chlorine:

The chloric(I) acid (\(\text{HClO}\)) is the active sanitising agent. It easily penetrates bacterial cell walls and kills them via rapid oxidation of essential proteins and enzymes.

⚠️ Chlorine in Sunlight

In the presence of bright sunlight (UV light), the reaction is different. Chlorine decomposes water to produce oxygen and hydrochloric acid instead:

\[ 2\text{Cl}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \xrightarrow{\text{UV light}} 4\text{HCl}(\text{aq}) + \text{O}_2(\text{g}) \]

This is why outdoor swimming pools require frequent replenishment of chlorine compared to indoor pools.

Alternative: Sodium Chlorate(I)

An alternative method of chlorination is adding solid sodium chlorate(I) (\(\text{NaClO}\)), also known as bleach, to water. In aqueous solution, this salt dissolves and establishes an equilibrium that produces the same disinfecting chloric(I) acid:

\[ \text{NaClO}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{Na}^+(\text{aq}) + \text{OH}^-(\text{aq}) + \text{HClO}(\text{aq}) \]

In alkaline conditions, the equilibrium shifts to the left, reducing the concentration of active \(\text{HClO}\). Therefore, the pH of the water must be carefully monitored and kept slightly acidic to neutral.

4. Benefits and Risks of Water Chlorination

In the exam, you are expected to weigh the arguments for and against chlorination of public water supplies.

Benefits Risks
Kills pathogens: Eradicates bacteria, viruses, and parasites that cause fatal waterborne diseases like typhoid and cholera. Toxicity: Chlorine gas is highly toxic and corrosive. Safe transport and storage require strict safety protocols.
Residual protection: Provides continuous disinfection as water flows through the distribution pipes to homes. Carcinogenic by-products: Chlorine can react with naturally occurring organic matter in water to form chlorinated hydrocarbons, such as trihalomethanes (THMs). These are linked to increased risks of bladder and colon cancers.
Cost-effective: Extremely cheap and highly scalable compared to alternatives like ozone treatment or UV radiation. Palatability: High concentrations can give water an unpleasant chemical taste and smell.
✏️ Worked Example: Explaining Pre-acidification
A student is testing a solution for halide ions. They add silver nitrate solution directly to the sample and observe a white precipitate. Explain why this result is invalid and describe the steps they should take to ensure the presence of chloride ions.

Answer:

  • The result is invalid because other ions, such as carbonate (\(\text{CO}_3^{2-}\)) or hydroxide (\(\text{OH}^-\)), form white precipitates of silver carbonate (\(\text{Ag}_2\text{CO}_3\)) or silver hydroxide (\(\text{AgOH}\)) when silver nitrate is added.
  • To correct this, the student must first add a few drops of dilute nitric acid (\(\text{HNO}_3\)) to the sample. The acid reacts with carbonate ions to form carbon dioxide gas and water, removing them: \[ \text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \]
  • After acidification, adding silver nitrate will form a white precipitate only if chloride ions are present.
  • To confirm the precipitate is silver chloride, the student should add dilute ammonia solution. The precipitate should dissolve completely to form a clear, colourless solution.
✏️ Worked Example: Disproportionation Calculation
Assign oxidation states to chlorine in the following reaction and use them to explain why the reaction is a disproportionation reaction: \[ \text{Cl}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaClO}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \]

Answer:

  • The reactant \(\text{Cl}_2\) is in its elemental state, so the oxidation state of chlorine is 0.
  • In \(\text{NaCl}\), sodium has a fixed oxidation state of \(+1\), so chlorine has an oxidation state of -1. Chlorine has gained an electron (undergone reduction).
  • In \(\text{NaClO}\), sodium is \(+1\) and oxygen is \(-2\). For the compound to be neutral, the oxidation state of chlorine must be +1 (since \(+1 - 2 + 1 = 0\)). Chlorine has lost an electron (undergone oxidation).
  • Because the chlorine atoms starting in \(\text{Cl}_2\) (oxidation state 0) have been both reduced to \(-1\) (in \(\text{NaCl}\)) and oxidised to \(+1\) (in \(\text{NaClO}\)) within the same reaction, this is a disproportionation reaction.
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