A student wanted to detect if nitrate ions were present in a solution. They firstly added iron(II) sulfate to the solution before slowly adding concentrated sulfuric acid. They noticed the formation of two layers with a colored ring at the interface. If nitrate ions were present, what color should the ring have been?
The purpose of this reaction is to detect if nitrate ions were present. In other words, this is a test for nitrate ions. When we carry out this process and nitrate ions are present, a ring of a certain color emerges. To better understand the test and to determine what color the ring is, let’s dive in to the reactions that are taking place.
First, we have a mystery solution that we will assume contains nitrate ions. For simplicity’s sake, let’s say that it’s sodium nitrate, although any cation would work for this example. First, we add iron(II) sulfate to that solution. The other reactant is sulfuric acid. As it turns out, the products of this reaction are iron(III) sulfate, sodium sulfate, water, and nitric oxide. The equation can be balanced with the coefficients written here. The important products here are iron(III) ions, water, and nitric oxide.
The next steps of the reaction are a little more complicated, but there’s just something we have to be familiar with. Six water molecules attach to an iron at the center to form a complex ion [Fe(H2O)6]2+. The nitric oxide bumps one of the water molecules out of the ion, creating the ion [Fe(H2O)5(NO)]2+ and a water molecule. This is the key ion, the end product of this process. As it turns out, it has a distinct brown color. When we add iron(II) sulphate and sulfuric acid to a solution that does indeed contain nitrate ions, this is the ion that forms.
Since we can easily view the brown ring in the test tube when we perform this test in the laboratory, it’s a reliable test for the presence of nitrate ions. We even call this test the brown ring test. So in this situation, if nitrate ions were present, the ring should have been brown.