Chemistry Forum

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.

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