How to make your own 75% Lugol's Iodine

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Take a 1 liter beaker and add 459g (3.614396865595038 moles) Iodine crystals (I2)

Add 514mL distilled water (not sure how this was calculated, was either from my experience or the solubility of KI in water) 

 

Add 225g (3.614396865595038 moles) Potassium hydroxide (90%) (KOH)

 

Will get hot.  Mix with ceramic or glass until fully dissolved.  Should be pale yellow clear solution.  (KOH and I2 can be added to dry beaker and stirred before adding water, but in some cases the heat from adding water hydrating the KOH might crack the beaker, this is why KOH is now added after the water.)

Let white precipitate settle down and filter out (or pour out liquid and keep liquid).  This white precipitate is potassium iodate and isn't good to ingest.

Add 250g iodine crystals to the liquid.  Add more water as necessary and gently heat to get it to dissolve.

Vol to 1L with distilled water.

This is 75% Lugol's iodine.

 

This is based on the recipe of:

 

500g KI (3.01 moles -> 3.614396865595038 moles of I and KOH due to loss of KIO3)

250g I2 

1L water

To make 5 mol KI you need 6 mol of Iodine and 6 moles of potassium hydroxide

6 I + 6 KOH => KIO3 + 5 KI

KOH = 56.11 g/mole

I = 126.904.47 g/mole

KI =  166.0028 g/mole

EDIT:

 

I had this recipe before, not quite perfect:

 

456g I2

225g KOH