In the equation:
4KI + 2CuSO4 => 2CuI(precipitate) + I2(aq) + 2K2SO4(aq)
To end up with a pure I2 product it would be necessary to sublime the CuI+I2 (after fine filtering the I2 out of soln.
Can anyone shed light on the validity of this method (which avoids the sublimation step):
1Filter out the CuI precipitate as per usual
2Then add HCl to the remaining soln (ie: I2+ K2SO4) until the Ph is 1.
3Then add H2O2 and the I2 will crash out of solnAxt - 10-12-2004 at 09:22
Quote:
Originally posted by johnnyBbad
1Filter out the CuI precipitate as per usual
2Then add HCl to the remaining soln (ie: I2+ K2SO4) until the Ph is 1.
3Then add H2O2 and the I2 will crash out of soln
Why use copper?, it can be precipitated straight from KI solution:
2KI + 2HCl + H2O2 --> I2 + 2KCl + 2H2O
It's what I use.BromicAcid - 10-12-2004 at 09:40
This was the method we used in lab to titrimetrically determine the copper concentration in ore.
2Cu2+ + 5I- ---> 2CuI + I3-
Followed by titration of the triiodide anion with sodium thiosulfate. I never noticed (solid) elemental iodine formed, the solution just went a nasty
brown color, i.e. iodine formed complexed with the iodide in solution to form triiodide. I wouldn't call this a good procedure to iodine by a
long shot just look at the potential for losses, and CuI is fairly insoluble unless you have excess ammonia or potassium iodide in solution (more
losses) so you'd have a hard time getting anything back from that ...
Quote:
....incredibly unstable, risking the potential hazard of explosion and a horrible oder of rotten eggs and mayonaise that will sink into your house and
not leave for weeks.
What?neutrino - 10-12-2004 at 16:25
You certainly wouldn’t get any back if you used ammonia. NI<sub>3</sub>… How would KI affect this reaction?johnnyBbad - 11-12-2004 at 18:24
thanks AXt.. I had my wires crossed a little there..your clear and concise response has clarified something for me..beautifully
ing for me..beautifully