Hello everyone,
I am writing my Bachelor´s thesis about different methods of quantify the phosphates into fertiliers.
I have experimented the "yellow method" in order to quantify the phosphates (PO43-) amount in form of sodium dihydrogen phosphate dihydrate (NaH2PO4 . 2 H2O) with one UV-VIS spectrometer, in order to make a calibration function. However, I don´t know how the reagent solution reacts with the phosphates and respectively with the sodium dihydrogen phosphate dihydrate (NaH2PO4 . 2 H2O):
I made a reagent solution of:
- Ammoniumvanadat (NH4VO3)
- Ammoniumheptamolybdat (NH4)6Mo7O24 . 4 H20
- nitric acid (HNO3)(65%) diluted in ratio 1: 3
I have just find this reaction :
PO4-3+ 12 MoO2-4 + 3 NH4++ + 24H++ ó (NH4)3(P(Mo3O10)4 (gelb) + 12 H2O
but I would like to know what happens with the Ammoniumvanadat?
Thank you very much in advance for your help,
Aurélie
Hi Aurélie Zouaoui , as far as I know, ammonium molbydate alone is perfectly OK to be used as a reagent to quantify phosphates. As shown in your equation it forms the yellow heteropolymolybdate salt (NH4)3(P(Mo3O10)4 (ammonium phosphomolybdate). Thus I'm wondering why you added the ammonium vanadate at all.
Hey Frank, thank you very much for you answer.
Actually, I have proceeded according the instructions of the book " Matissek, R., Steiner, G., Fischer, M. (2010). Lebensmittelanalytik, 4., vollständig überarbeitete Springer" and I am also wondering the reason why the vanadate was necessary. Unfortunately, I cannot try the experiment again without vanadate, because of the corona crisis I don´t have any access to the laboratory anymore...
However, I would like to find out how it is working by the other phosphate detection method, called "blue-method".
Maybe could you help me to find out how the equation looks like?
The reactants are :
- Phosphates
- Ammoniummolybdate
- L-ascorbic acid
- sulphuric acid
- destilled water
One of the formed product is phosphate-molybdenum blue complex
thank you very much for your help
Sincerely,
Aurélie .
Hi Aurélie Zouaoui, in this case the L-ascorbic acid acts as reducing agent. It reduces the Mo(VI) to a somewhat lower oxidation state between +5 and +6 (see Wikipedia article "Molybdänblau"). In both cases the addition of ammonium vanadate is not necessary.
https://de.wikipedia.org/wiki/Molybdänblau
https://www.bsz.uni-rostock.de/storages/uni-rostock/Alle_MNF/Bio_Zingt/Dokumente/SOPs/An_Phosp.pdf
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