Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
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sulphuric acid : H2SO4
sodium hydroxide : NaOH
hydrogen peroxide : H2O2
thiourea : NH2 CS NH2
Class:
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elem=C,N,S
precipitation
redox
Summary:
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Thiourea is oxidized in acidic solution to formamidine disulfide, which exists
in solution as an acid salt. When a strong base is added, then the free base is
formed, but this at once decomposes. One of the decomposition products is
sulphur.
When excess oxidizer is used and the pH is increased strongly, then the sulphur
dissolves, giving rise to formation of polysulfides.
Description:
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Dissolve some thiourea in water and add a little dilute sulphuric acid: A clear
and colorless solution is obtained.
Add a slight excess amount of hydrogen peroxide (10% by weight): The liquid
becomes warm and remains colorless and clear.
Add a small quantity of a solution of sodium hydroxide: At once a very pale
yellow precipitate is formed. On shaking this precipitate spreads over the
entire liquid. The liquid does not become clear anymore.
Add more solution of sodium hydroxide: The precipitate becomes denser/thicker.
Add even more sodium hydroxide, such that the liquid becomes strongly alkaline:
The liquid slowly turns intense yellow and slowly the precipitate dissolves.
After a few tens of seconds the liquid is clear and bright yellow. A few
minutes later, the liquid still is clear, but the yellow color has faded
somewhat (it still is quite intense though).
Add a few drops of dilute sulphuric acid: A very pale yellow precipitate is
formed, which on shaking redissolves again and the liquid becomes intense
yellow again. More intense than it was before adding the acid.
Add much more dilute sulphuric acid: A thick white precipitate is formed and a
strong smell of hydrogen sulfide (rotten eggs) is produced.
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In this experiment, first the thiourea is oxidized to formamidine disulfide in
the acidic solution. Formamidine disulfide is protonated at its two NH2-groups.
When excess hydroxide is added, then the formamidine disulfide decomposes and
one of the products is sulphur. On addition of more hydroxide, such that the
solution becomes strongly alkaline, the sulphur dissolves and polysulfide is
formed in the solution (this gives the strong yellow color). When the solution
is acidified again, then the polysulfide