I can't seem to figure out a reaction for CuSO4*5H2O + (NH4)2S2O8 in the presence of pyridine to make [Cu(py)4]S2O8. Where should the waters go? Should the SO4 join with the ammonium?
Same as for AgNO3 and (NH4)2S2O8 in the presence of pyridine to make [Ag(py)4]S2O8.
Please, please, pleaaase help me out if you're up. I'll probably be up for another 3-4 hours.
URGENT CuSO4*5H2O + (NH4)2S2O8?????
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- ChenBeier
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In aqueous solution you have ions present, what means Cu2+, NH4+ SO4 2-, S2O8 2-
Theoretically you can write Cu2+ + 4 py + S2O8 2- => Cu(py)4S2O8
Ammonium and Sulfate is remaining.
CuSO4*5H2O + (NH4)2S2O8 + 4 py => Cu(py)4S2O8 + (NH4)2SO4 + 5 H2O
For silver its similar.
2 AgNO3 + (NH4)2S2O8 +4 py => Ag2(Py)4S2O8 + 2 NH4NO3
Theoretically you can write Cu2+ + 4 py + S2O8 2- => Cu(py)4S2O8
Ammonium and Sulfate is remaining.
CuSO4*5H2O + (NH4)2S2O8 + 4 py => Cu(py)4S2O8 + (NH4)2SO4 + 5 H2O
For silver its similar.
2 AgNO3 + (NH4)2S2O8 +4 py => Ag2(Py)4S2O8 + 2 NH4NO3