Preparation of iodine solution

How to prepare $\mathrm{0.1M}$ $\ce{I\_2}$ solution.  

I have found in books that it should be prepared in $\ce{KI}$ solution but none of them explains why? Please explain in detail not just the procedure.




Since $\ce{ I\_2}$ is a non-polar covalent molecule, it does not ionize in water. It cannot also be soluble in the polar water.


$\ce{KI}$, which is a polar, ionic compound, will ionize and dissolve in water. When KI dissolves in water, it ionizes to $\ce{K^+}$ and $\ce{I^-}$.


The $\ce{I^-}$ will react with $\ce{ I\_2}$ to form the complex ion $\ce{I\_3^-}$. $\ce{I\_3^-}$ being negatively charged will dissolve in water.


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