Question

Why is iodine almost always used in a solution containing excess ${\mathbf{l}}^{\mathbf{-}}$?

Short answer

Iodine is always used in a solution with excess ${\mathrm{l}}^{-}$ because pure${\mathrm{l}}_2$ is not a polar substance and it cannot be dissolved in water.

Step-by-step solution

Properties of Iodine.

• Iodine is a chemical element with atomic number 53 and the symbol l .
• At typical conditions, it exists as a semi-lustrous, non-metallic solid that melts to produce a deep violet liquid at $114°$C and boils to form a violet gas at $184°\mathrm{C}$
• Iodine is found in a variety of oxidation states, including iodide$\left({\mathrm{I}}^{-}\right)$ , iodate $\left({{\mathrm{IO}}_3}^{-}\right)$, and periodate anions.
• It is the heaviest nutrient in terms of necessary minerals.

To determine the iodine always used in solution containing  

• lodine is always used in a solution with excess${\mathrm{I}}^{-}$because pure${\mathrm{I}}_2$is not a polar substance and it cannot be dissolved in water.
• Because of that, iodide solution(for example, Kl) must be added to iodine to increase its solubility,
• The following reactions occurs:
• ${\mathrm{I}}_2\left(\mathrm{s}\right)+{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{I}}_3^{-}\left(\mathrm{aq}\right)$
• In this reaction triodide is formed by a complexation reaction between iodide and iodine and the solubility of ${\mathrm{I}}_2$ is increased.