Testing for Halide Ions

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This page discusses the tests for halide ions (fluoride, chloride, bromide and iodide) using silver nitrate and ammonia.

Using silver nitrate solution

This test is carried out in a solution of halide ions. The solution is acidified by adding dilute nitric acid. The nitric acid reacts with, and removes, other ions that might also form precipitates with silver nitrate. Silver nitrate solution is then added, and the halide can be identified from the following products:

ion present	observation
F^-	no precipitate
Cl^-	white precipitate
Br^-	very pale cream precipitate
I^-	very pale yellow precipitate

The chloride, bromide and iodide precipitates are shown in the photograph:

The chloride precipitate is easily identified, but the other two are quite similar to each other. They can only be differentiated in a side-by-side comparison. All the precipitates change color if they are exposed to light, taking on gray or purple tints. The absence of a precipitate with fluoride ions is unhelpful unless it is known that a halogen is present; otherwise, it indicates that there is no chloride, bromide, or iodide.

The chemistry of the test

The precipitates are insoluble silver halides: silver chloride, silver bromide or silver iodide. The formation of these is illustrated in the following equations:

\[ Ag^+_{aq} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} \nonumber \]

\[ Ag^+_{aq} + Br^-_{(aq)} \rightarrow AgBr_{(s)} \nonumber \]

\[ Ag^+_{aq} + I^-_{(aq)} \rightarrow AgI_{(s)} \nonumber \]

Silver fluoride is soluble, so no precipitate is formed.

\[ Ag^+_{aq} + F^-_{(aq)} \rightarrow Ag^+_{aq} + F^-_{(aq)} \nonumber \]

Confirming the precipitate using ammonia solution

Ammonia solution is added to the precipitates.

original precipitate	Observation
AgCl	precipitate dissolves to give a colorless solution
AgBr	precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colorless solution
AgI	precipitate is insoluble in ammonia solution of any concentration

There are no absolutely insoluble ionic compounds. A precipitate forms if the concentrations of the ions in solution in water exceed a certain value, unique to every compound. This value is known as the solubility product. For the silver halides, the solubility product is given by the expression:

\[ K_{sp} = [Ag^+][X^-] \nonumber \]

The square brackets indicate molar concentrations, with units of mol L^-1.

If the product of the concentrations of ions is less than the solubility product, no precipitate is formed.
If the product of the concentrations exceeds this value, a precipitate is formed.

Essentially, the product of the ionic concentrations is never greater than the solubility product value. Enough solid is always precipitated to lower the ionic product to the solubility product. The table below lists solubility products from silver chloride to silver iodide (a solubility product for silver fluoride cannot be reported because it is too soluble).

	K_sp(mol²dm^-6)
AgCl	1.8 x 10^-10
AgBr	7.7 x 10^-13
AgI	8.3 x 10^-17

The compounds are all quite insoluble, but become even less so down the group.

The purpose of ammonia

The ammonia combines with silver ions to produce a complex ion called the diamminesilver(I) ion, [Ag(NH₃)₂]⁺. This is a reversible reaction, but the complex is very stable, and the position of equilibrium lies well to the right. The equation for this reaction is given below:

A solution in contact with one of the silver halide precipitates contains a very small concentration of dissolved silver ions. The effect of adding the ammonia is to lower this concentration still further. If the adjusted silver ion concentration multiplied by the halide ion concentration is less than the solubility product, some precipitate dissolves to restore equilibrium.

This occurs with silver chloride, and with silver bromide if the ammonia is concentrated. The more concentrated ammonia pushes the equilibrium even further to the right, lowering the silver ion concentration even more.

The silver iodide is so insoluble that ammonia cannot lower the silver ion concentration enough for the precipitate to dissolve.

An alternative test using concentrated sulfuric acid

Adding concentrated sulfuric acid to a solid sample of one of the halides gives the following results:

ion present	observation
F^-	steamy acidic fumes (of HF)
Cl^-	steamy acidic fumes (of HCl)
Br^-	steamy acidic fumes (of HBr) contaminated with brown bromine vapor
I^-	some HI fumes with large amounts of purple iodine vapor and a red compound in the reaction vessel

The only possible confusion is between a fluoride and a chloride—they behave identically under these conditions. They can be distinguished by dissolving the original solid in water and then testing with silver nitrate solution. The chloride gives a white precipitate; the fluoride produces none.

Contributors and Attributions

Jim Clark (Chemguide.co.uk)