Why in solutions of NaI in some solvent (formamide, N-methilformamide or N-methilacetamide pure on in aqueous mixtures) forms a yellowish color with the passage of time? I have tried to make solutions by removing the air (by vacuum or bubbling through nitrogen or argon) and yellow is still forming. This doesn't happen with the NaF, NaCl and NaBr.
I think it is due to the spontaneous reduction of iodide (I-) to Iodine (I2), but I cannot say why it happens. Did you measure the redox potential in the solution?.
Try to add a small amount of any oxidant, probably the yellow color will not appear again.
I- ions could be oxidized I2. and then I- make a complex with I2 to form a complex (I-3) which is called tri-iodide ion.and this complex is yellowish in colour. this exist in the solution as an equilibrium as follows
(I-) + (I2) = (I-3)
Iodides are slowly oxidized (not reduced!) by atmospheric oxygen to molecular iodine. Its solubility is increased when forming the triiodide ion as above. Water (moisture) facilitates the oxidation process. The iodide solution should be stored at low T and low humidity conditions, with exclusion of air.
Iodine may also form various colored charge transfer complexes (for example with pyridine).
Iodide can also produce iodine by photooxidation, initially converting to atomic I and solvated electron. Hence, the iodide solution may also be stored in the dark.
In addition to des-airear (with vacuum or argon bubbling) I kept the dissolutions in the dark. Already had read something about the I3 - but (I have some articles about) but do not justify increasing of the yellow colour with the passage of time and less if it is non-aqueous solvents!
KI solutions can form I2 by the light or the air. Add few crystals of Na2S2O3 (or others reducing compounds like NaN3) for keep stable the solution (If the thiosulphate doesn't give any interference for your further use)
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