Practical investigations

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I have used colorimetry to find the concentration of manganese in tea by comparing the results with different concentrations of potassium manganate(VII) solution. I used the following method to extract the manganese from the tea:
 
Strongly heat the dry tea to give a white ash.
Add 15 cm³ distilled water, 3 cm³ concentrated sulphuric acid, 4 cm³ 85% phosphoric acid, 8 cm³ concentrated nitric acid, and boil.
Then add 50 cm³ 0.01M silver nitrate solution and 2.5g ammonium peroxodisulphate, and boil to give purple solution.
 
I wanted to know what reactions are taking place when each different chemical is added. Thanks.

Igloo writes ...
 
I suspect that the manganese is present in the form of Mn²⁺ ions in the tea.
These ions need to be leeched out of the leaves, first by burning off all the cellulose and volatile materials. This burning may well convert some of the Mn²⁺ ions to MnO₂ in which manganese is in its +4 oxidation state. Secondly the boiling up with the mixture of acids is likely to ensure that all the manganese has been oxidised to MnO₂ or even manganate(VI), MnO₄^2- ions.
 
Finally, the silver ions in the silver nitrate, and notably the peroxodisulphate ions in the ammonium peroxodisulphate, will oxidise the manganese through to its seventh oxidation state, i.e. to purple manganate(VII) ions, MnO₄^-. Peroxodisulphate ions have an even higher redox potential than manganate(VII) ions so this ensures that the conversion occurs.
 
Fortunately you haven’t asked me for equations, so I shan’t even try!
 
I am certain that a great many different redox reactions are occurring at stage 2, for example, but I’m sure that I have given you an accurate, but rough gist of what’s going on.

Risk assessment
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updated: 20 December 2007