Water and purity is a very big environmental concern. Pure water contains no dissolved ions, and many tests have been devised to determine the presence of ions in a sample of water. In this lesson, we will learn how to test for chloride, bromide, and iodide ions. If we are working with unknown solids, they must first be dissolved in water. The resulting solution must then be acidified with dilute nitric acid. This removes any impurities that may be present to give a clear result. Then add a few drops of aqueous silver nitrate. Silver halide precipitates are formed, and each has a characteristic color. If chloride ions are present, the resulting silver chloride precipitate is white. If bromide ions are present, the resulting silver bromide precipitate is cream. If iodide ions are present, the resulting silver iodide precipitate is yellow. It may be difficult to distinguish between the white silver chloride precipitate and cream silver bromide precipitate. Add dilute aqueous ammonia. Add aqueous ammonia dissolves silver chloride but not silver bromide. Therefore, this test is used to distinguish between silver chloride and silver bromide. Concentrated ammonia dissolves both silver chloride and silver bromide. Where iodide dissolves in neither dilute nor concentrated ammonia. Write down the ionic equations for the precipitation reactions that we have learned about. Resume the videos once you are done. The equation are as follow. Remember that since the silver halides are precipitates, the subscript is S in round brackets. Halide ions can be detected through the addition of aqueous silver nitrate, which forms a silver halide precipitate. To conclude, silver chloride is white and dissolves in dilute aqueous ammonia. Silver bromide is cream and only dissolves in concentrated ammonia. Silver iodide is yellow and does not dissolve in either dilute or concentrated ammonia.