# Explanation of Iodine Equilibrium in immiscible solvents

• kirsten_2009
In summary, the equilibrium of I2 and its distribution in two immiscible liquids (varsol and aqueous potassium iodide) is determined by the solubility of iodine in both layers. In varsol, iodine is present as I- ions, giving a pink color, while in the aqueous layer, it is present as I3- ions, giving a yellow color. When more iodine or KI is added, the equilibrium shifts to redistribute the iodine between the layers. This is seen in Scenario 1 and 2, where adding more I2/KI or KI affects the color of the layers. Changing the concentration of any of the reagents will also affect the distribution of iodine between the
kirsten_2009

## Homework Statement

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.

## Homework Equations

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).

kirsten_2009 said:
Iodine is in varsol it's in the form of I- ions I2 molecules = pink color.
Other than that little correction (you've got it that way elsewhere) I'd say you've understood the problem.

Great, thank-you! And as a follow up to all this and as a conclusion...in the I2 (aq) + I- (aq) <----> I3- (aq) equation if any of the reagent's concentration was changed...this would affect the iodine distribution between the two liquids, correct? But, let's say I3- was increased would that mean that the aqueous layer would be greater or more yellow because there are more I3- in the aqueous layer or would it means that because I3- was increased the reaction would shift to the left and more I2 and I- would be formed in the varsol layer thus making the varsol layer greater and more pink?

kirsten_2009 said:
would that mean
kirsten_2009 said:
or would it
"Yes." Both processes are going to affect the "partition" of iodine between the phases, hence, the study of "partition coefficients."

## 1. What is the iodine equilibrium in immiscible solvents?

The iodine equilibrium in immiscible solvents refers to the equilibrium between iodine molecules and iodide ions in two solvents that are unable to mix together. This equilibrium is important in understanding the behavior of iodine in various chemical reactions and in the production of iodine-based compounds.

## 2. How does the iodine equilibrium work in immiscible solvents?

The iodine equilibrium in immiscible solvents is based on the principles of solubility and polarity. Iodine is more soluble in nonpolar solvents, such as hexane, while iodide ions are more soluble in polar solvents, such as water. The equilibrium is reached when the concentrations of iodine and iodide ions in each solvent are in balance.

## 3. What factors affect the iodine equilibrium in immiscible solvents?

The factors that affect the iodine equilibrium in immiscible solvents include the relative solubility of iodine and iodide ions in each solvent, the temperature, and the presence of other chemicals that may interact with iodine or iodide ions.

## 4. Why is the iodine equilibrium in immiscible solvents important in chemistry?

The iodine equilibrium in immiscible solvents is important in chemistry because it allows for the separation and purification of iodine from other substances. It also plays a key role in many chemical reactions involving iodine, as the equilibrium concentration of iodine can affect the rate and yield of the reaction.

## 5. How is the iodine equilibrium in immiscible solvents measured?

The iodine equilibrium in immiscible solvents can be measured using various techniques, such as spectrophotometry or titration. These methods involve measuring the concentration of iodine or iodide ions in each solvent and calculating the equilibrium constant, which is a measure of how strongly the equilibrium favors the formation of iodine or iodide ions.

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