I add 5 drops Potassium Iodide (KI) to 5 drops lead nitrate (PbNO3). The solution is centrifuged and the aqueous layer is discarded. I then add 5 drops of nitric acid (HNO3), 5 drops of hydrogen peroxide and 10 drops of dichloroethane (C2H4Cl2) to the precipitate.
The solution is shaken and lightly heated resulting in a clear aqueous solution with a deep red precipitate.
What anions are present in the precipitate? How do you know?
Appreciate the help.
From your description, I think you have formed red lead oxide (Pb3O4), The oxide ion is present in the ppt.
To determine the anions present in the precipitate formed after adding nitric acid, hydrogen peroxide, and dichloroethane to the solution, we need to analyze the chemical reactions that occur during the process.
1. Firstly, when you add 5 drops of potassium iodide (KI) to the 5 drops of lead nitrate (Pb(NO3)2), a double displacement reaction takes place. The lead cation (Pb2+) from lead nitrate reacts with iodide anion (I-) from potassium iodide to form lead iodide (PbI2) precipitate according to the following equation:
Pb(NO3)2 (aq) + 2 KI (aq) -> PbI2 (s) + 2 KNO3 (aq)
2. When you centrifuge the mixture, you discard the aqueous layer. This step helps in separating the solid precipitate (PbI2) from the remaining solution.
3. Next, you add 5 drops of nitric acid (HNO3). Nitric acid acts as an acid to fully dissolve any remaining lead iodide precipitate from the previous step.
4. Then, you add 5 drops of hydrogen peroxide (H2O2). Hydrogen peroxide serves as an oxidizing agent, oxidizing any remaining iodide (I-) to iodine (I2).
5. Finally, you add 10 drops of dichloroethane (C2H4Cl2) to the mixture and shake it. Dichloroethane is a nonpolar solvent that helps extract the iodine (I2) from the aqueous solution.
The deep red color in the resulting aqueous solution indicates the presence of iodine (I2). Therefore, the anion present in the precipitate is iodide (I-), which was oxidized to iodine (I2) during the reaction.