Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
-----------------
sulphuric acid : H2SO4
sodium chlorate : NaClO3
potassium iodide : KI
potassium bromide : KBr
potassium bromate : KBrO3
Class:
------
elem=Br,I,Cl
redox
Summary:
--------
Bromate oxidizes iodine to iodate, itself being converted to bromine.
Chlorate does not oxidize bromine to bromate, itself being converted to
chlorine (at least not quickly).
Description:
------------
Sequence 1:
------------
Prepare a small amount of bromine, by adding a slight excess amount of
potassium bromate to some potassium bromide and adding dilute sulphuric
acid (2 mol/l) and add a large excess amount of sodium chlorate: The sodium
chlorate quickly dissolves and the liquid becomes a little turbid (probably
some sodium sulfate and/or potassium chlorate is formed, which makes the
liquid slightly turbid). The color of the bromine does not disappear.
Heat the liquid to appr. 60C: The liquid becomes completely clear and orange/
red. The vapour of bromine becomes thicker above the liquid. It does not
disappear.
Sequence 2:
------------
Add some potassium bromate to dilute sulphuric acid (2 mol/l): The solid only
dissolves very slowly. The liquid remains completely clear and colorless.
Add a small amount of solid potassium iodide to the liquid: The solid dissolves
and small black particles of solid iodine are formed immediately. The iodine
quickly sinks to bottom. The liquid above the iodine is yellow/orange.
Heat to appr. 60C: All solids dissolve, including the iodine. The iodine does
not make the liquid brown, the liquid becomes darker orange/red and a brown
vapour of bromine appears above the liquid. Finally, the liquid is completely
clear, no solids are remaining and a fairly thick vapour of bromine is above
the liquid. All iodine is converted, the only reasonable thing to assume is
that it is converted to iodate.