thebean - 5-1-2014 at 16:57
Today I needed to clean out a vile that contained a small amount of iodine. I added hot water knowing that I2 is slightly soluble in water and would
give a nice orange/red color. I added some NaOH in the hopes of producing water soluble NaIO3 which it did, evidenced by the loss of color in the
solution and the other I2 dissolving. I thought I might have added too much NaOH so I used a small amount of 6% AcOH to neutralize any remaining NaOH.
A small amount of bubbling occurred and then the reddish orange color from before returned. I understand how the addition of the base affected the
solution but I don't fully get the part with the AcOH because no I2 precipitated.
eidolonicaurum - 6-1-2014 at 04:43
Iodine reacts with water like this:
I2 + H2O <-> HI + HIO
By adding the alkali, you pushed the equilibrium all the way to the right. This removed any free iodine, an therefore the colour. By acidifying the
solution again, you reestablished the equilibrium, causing the colour to come back.