I'm studying the equilibrium of I2 and its distribution in two immiscible liquids (varsol and aqueous potassium iodide). Since I2 is non polar; it will "prefer" to be in varsol but since KI (aq) is an ionic salt solution, iodine will also dissolve to give the I3- ion. When iodine is present in varsol = pink color and when in KI = yellow.
I think I understand what happens but I'm a little confused as to what would be observed in the following situations, any clarification would be really appreciated.
Scenario 1: 2 mL iodine/potassium iodide (aq) + 2 mL of varsol....."shake"...then add saturated solution of I2/KI .....what would be observed? What is happening on a molecular level?
Scenario 2: 2 mL iodine/potassium iodide (aq) + 2 mL of varsol....."shake"...then add saturated KI...same questions.
I2 (aq) + I- <---> I3- (aq)
I2 (aq) <---> I2 (varsol)
The Attempt at a Solution
So this is my understanding of the situation, please correct me if I'm wrong. Iodine will be present in both organic and aqueous layer because its soluble in both. When Iodine is in varsol it's in the form of I- ions = pink color. When iodine is present in the aqueous KI solution its in the form of I3- ions = yellow color. So initial solutions should appear with pink varsol layer on top and yellow aqueous layer at bottom.
Scenario 1: When saturated solution of I2/KI is added I would think that nothing (no change) is observed and that the layers continue to be pink on top and yellow at bottom because more iodine is being added but also more KI so the iodine will just redistribute itself in the two layers until it's reached equilibrium. Iodine can just have more I3- since it has more solvent now to do so in (equilibrium would shift right).
Scenario 2: When saturated KI solution is added I would think that the pink layer would become lighter and the yellow layer more yellow since there is more KI for iodine to dissolve in and so more I3- ions would be formed (equilibrium would shift right).