How to make Lugol's Iodine

Lugol's solution consists of
1. 5 g iodine (I2) and
2. 10 g potassium iodide (KI) mixed with
3. 85 ml distilled water, to make a brown solution with a total iodine content of 130 mg/mL.

To make the original version, we use what amounts to a 3% (2.91%) formula (iodine and iodine compound). The formula is simple: 80 gm. of potassium iodine (KI) is added to 4,000 ml of distilled water (roughly 4,000 gm). After fully dissolving, 40 gm. of iodine crystal is added. After the iodine crystal fully dissolves (which takes several hours), the product is packaged and stored.
To make the newer 7% Lugol's version, using the same 4,000 ml. of distilled water, the amounts of iodine use are 200 gm. of KI, and 100 gm. of iodine crystal. For 15% Lugol's, to 4,000 ml. of distilled water we dissolve 480 gm. of KI and 240 gm. of iodine crystal.

Potassium iodide renders the elementary iodine soluble in water through the formation of the triiodide (I3−) ion. It is not to be confused with tincture of iodine solutions, which consist of elemental iodine, and iodide salts dissolved in water and alcohol. Lugol's solution contains no alcohol.