The formation of a white precipitate which becomes yellow and then black
name: The formation of silver iodate followed by reduction to iodide. Further, the free iodine is released which forms an iodine-starch complex.
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AgNO3 + KIO3 → AgIO3 + KNO3
2 AgIO3 + 6 Na2S2O5 → 2 AgI + 6 Na2SO4 + 6 SO2
2 AgI → 2 Ag + I2
- First, it is necessary to prepare three solutions, in large beakers. Achieve a complete dissolution of the components in each cup.
1) Add 20 ml of distilled water to 0.7 g of sodium metabisulfite. Measure volumes using measuring cylinders.
2) Pour 0.5 g of potassium iodate into a beaker and add 35 ml of water. Due to the low solubility in water, potassium iodate solution may be heated on a hot plate. Add a small amount of pre-cooked starch paste solution. Add 120 more ml of water.
3) Take a sample of silver nitrate (0.17 g). Then add 40 ml of distilled water.
- To perform the reaction mix all three solutions. During the reaction, the mixture must be regularly stirred with a glass rod.
The appearance of the white precipitate at the start is due to an ordinary ion exchange reaction between silver nitrate and potassium iodate which leads to the formation of silver iodate.
Sodium metabisulfite has reducing properties, which allow it to reduce iodate to iodide. Thus, the solution becomes yellow.
Silver iodide is an unstable compound and may decompose to simple substances. (This property has been used in photography.) Isolated free iodine forms a stable complex with starch that has a black color.