Phosphate can be analyzed spectrophotometrically by conversion to 12-molybdophos
ID: 958206 • Letter: P
Question
Phosphate can be analyzed spectrophotometrically by conversion to 12-molybdophosphoric acid, H3PO4(MoO3)12. Sodium molybdate, Na2MoO4*2H2O is added to a phosphate containing solution, along with hydrazine sulfate, H3NNH3(2+)*SO4(2-), producing a blue colored solution.Write a balanced equation for the reaction forming 12-molybdophosphoric acid.
I am very confused on how these chemicals all relate to one another as a reaction. I feel that the question is unclear, especially as to what the blue colored solution might be? Phosphate can be analyzed spectrophotometrically by conversion to 12-molybdophosphoric acid, H3PO4(MoO3)12. Sodium molybdate, Na2MoO4*2H2O is added to a phosphate containing solution, along with hydrazine sulfate, H3NNH3(2+)*SO4(2-), producing a blue colored solution.
Write a balanced equation for the reaction forming 12-molybdophosphoric acid.
I am very confused on how these chemicals all relate to one another as a reaction. I feel that the question is unclear, especially as to what the blue colored solution might be?
Write a balanced equation for the reaction forming 12-molybdophosphoric acid.
I am very confused on how these chemicals all relate to one another as a reaction. I feel that the question is unclear, especially as to what the blue colored solution might be?
Explanation / Answer
There are two reaction: non-redox formation of phosphomolybdic acid H3PMo12O40 in acidic medium((HCl) followed by reduction to Mo5O15 after addition of SnCl2 to this mixture.
(NH4)6Mo7O24 + Na2HPO4 + HCl = H3PMo12O40 + NH4Cl + NaCl
H3PMo12O40 +SnCl2 + HCl = Mo5O14 + SnCl4 + H3PO4 + H2O
Try to balance above reactions