Iodine is mainly found in water in the trioxide ion, I3 which is formed by the combination of molecular iodine, I2 and Iodide, I-, according to the equation:
I2(aq) +I-(aq) <----> I3(aq)
QUESTION: how does the system in the equilibrium seem to respond to the changes applied to it with respect to the distribution of iodine considering that I2 will partition between the two phases I2(aq) <--> I2 (varsol)?
I've never done this; however, if I2 partitions more easily to the varsol(which it will do) it would appear to me that it increases I2 in varsol which reduces I2 in the aqueous phase and that should shift the I2==>I3- equilibrium to the left.
By the way, you meant to call I3^- triiodide in your opening remarks.
Thank you so much
To understand how the system in equilibrium responds to changes in terms of iodine distribution, we need to consider Le Chatelier's principle. According to this principle, when a system in equilibrium is subjected to a change, it will respond in a way that partially offsets the effect of that change.
In this case, the system is in equilibrium between the two phases: I2(aq) and I2(varsol), where varsol is an organic solvent. The concentration of iodine in each phase determines the distribution of iodine between them. Now, let's consider a few possible changes and how the system responds:
1. Increase in I2(aq):
- According to Le Chatelier's principle, the system will respond by shifting the equilibrium to the left, i.e., towards the I2(aq) phase.
- As a result, more I2 will dissolve in the aqueous phase, increasing the concentration of I2(aq) and reducing the concentration of I2(varsol).
2. Increase in I2(varsol):
- Similarly, if the concentration of I2(varsol) is increased, the system will respond by shifting the equilibrium to the right, towards the I2(varsol) phase.
- This will cause more I2 to partition into varsol, increasing the concentration of I2(varsol) and reducing the concentration of I2(aq).
3. Decrease in I2(aq) or I2(varsol):
- A decrease in the concentration of I2 in either phase will cause the system to respond by shifting the equilibrium in the opposite direction - towards the phase with higher I2 concentration.
- The system will try to compensate for the decrease by shifting the equilibrium to increase the concentration of I2 in the depleted phase.
It's important to note that the distribution of iodine will depend on the solubility and affinity of I2 towards each phase. Without any additional information about the solubilities of I2(aq) and I2(varsol), we cannot predict the exact distribution of iodine under various conditions. These changes will simply affect the equilibrium position and hence influence the iodine distribution to some extent.
To determine the actual distribution of iodine, one would need to measure or know the solubilities of I2 in water and varsol and use appropriate calculations or experiments to determine the concentrations in each phase at the given conditions.