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.


1 Answer 1


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.

  • $\begingroup$ And in what amount KI should be taken? Does the amount matter in preparing a solution of particular concentration? $\endgroup$ Dec 6, 2014 at 1:28
  • 1
    $\begingroup$ Ingredients 1. Potassium iodide: 10 g 2. Distilled water: 100 ml 3. Iodine crystals: 5 g. Preparation: A. Dissolve 10 g potassium iodide in 100 ml of distilled water. B. Slowly add 5 g iodine crystals, while shaking. C. Filter and store in a tightly stoppered brown bottle. This solution has the name of Lugol's iodine solution. As you can see, it has 0.0197 mol of iodine and 0.06 mol of potassium iodide. $\endgroup$ Dec 6, 2014 at 14:08

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