There is Lugol's 5 % solution with 5 g of pure iodine and 7.5 g of potassium iodide. In one ml there should be 12.5 mg of iodine in sum.

But there are suggestions telling that two drops of this solution contains 12.5 mg of iodine and the drops I seen used there are no way 1 ml. Is it a mistake? Why they don't use measurement in ml rather than drops to be precise?

  • $\begingroup$ It's true, 2 drops of Lugol's 5% contain 12.5 mg of iodine/iodide, see this link for the calculation. You cannot just add up the amounts of iodine and iodide, they are not chemically identical. Also, a well-calibrated dropper is much more precise than measuring 1 ml in a syringe. $\endgroup$
    – Don_S
    Commented Mar 1, 2017 at 12:21
  • $\begingroup$ 2 drops would be about 0.1 mL, 20 drops would be about 1mL. While this can be a surprisingly accurate means of measuring aqueous solutions, I disagree with the previous comment; a properly calibrated 1 mL syringe will give you a more accurate and precise delivery of 1 mL than the dropper technique. Depending on your needs though, the dropper method might be sufficient. $\endgroup$
    – airhuff
    Commented Mar 1, 2017 at 12:56
  • $\begingroup$ @Don_S So are you saying in 1 ml there is no 12.5 mg of iodine in sum? 5 % solution in this case means 5 g pure iodine and 7.5 g of potassium iodide so logically in 1 ml of this solution there should be 5 mg of pure iodine and 7.5 mg of potassium iodide? Is not it correct? $\endgroup$
    – Sofiko
    Commented Mar 1, 2017 at 18:23
  • $\begingroup$ @airhuff there is a big difference between two drops and twenty drops. This solution is intended to be used internally and droppers are comfortable for it but thought it was imprecise. $\endgroup$
    – Sofiko
    Commented Mar 1, 2017 at 18:26
  • $\begingroup$ @airhuff, my intention was that a regular scaled syringe of 5 or 10 mL (even if scaled very precisely) would still not be a match to a dropper, such as the opening in liquid iron supplements for babies, where the opening is designed to give the same volume in every drop - sure, they also fail sometimes for a bunch of reasons, but compared to the syringe, where you have to pull and push manually and you depend on bringing the end of the piston exactly to the scale by sight alone, the error rate of a dropper is probably lower. $\endgroup$
    – Don_S
    Commented Mar 1, 2017 at 18:32

2 Answers 2


A Lugol's iodine solution is an aqueous solution of elemental iodine and potassium iodide. This solution may be prepared in different strengths, typically $2{-}15~\%$, while the strength is determined by the content of elemental iodine (see this table from the Wikipedia article, but note that the first column's title is a bit misleading — the leftmost column lists the different strengths of Lugol's solution, and the other columns list the amount of the each constituent (iodine, iodide, total) in a solution of that strength).

Now, this is in essence an iodine solution, but it is prepared from iodine and potassium iodide due to solubility issues of elemental iodine. See this answer on Chemistry.SE to the question Preparation of iodine solution for a simple and concise explanation.

For the calculation of the amount of iodine and potassium iodide in solution, let us take $5~\%$ Lugol's solution, which is prepared from 5 grams of iodine and 7.5 grams of potassium iodide in $\pu{100 mL}$.

If we have $\pu{5 g}\ (\pu{5000 mg})$ of iodine and $\pu{7.5 g}\ (\pu{7500 mg})$ of potassium iodide in $\pu{100 mL}$, then after dividing by $100$, we have $\pu{50 mg}$ of iodine and $\pu{75 mg}$ of potassium iodide in $\pu{1 mL}$.

Since the commonly accepted volume of one drop is $1/20\ (5~\%)$ of $\pu{1 mL}$, then by multiplying $\pu{50 mg}$ and $\pu{75 mg}$ by $0.05$, we get $\pu{2.5 mg}$ of iodine and $\pu{3.75 mg}$ of potassium iodide in one drop of the solution.

For better clarity, here is the calculation again for I2 content:

$$\frac{\pu{5 g}}{\pu{100 ml}}\cdot \frac{\pu{1000 mg}}{\pu{1 g}} = \frac{\pu{50 mg}}{\pu{1 mL}} \\ \Longrightarrow \frac{\pu{50 mg}}{\pu{1 mL}}\cdot \frac{\pu{1 mL}}{\pu{20 drops}} = \frac{\pu{2.5 mg}}{\pu{1 drop}}$$

Do the same for $\pu{7.5 g}$ of potassium iodide to get $\pu{3.75 mg/drop}$.

A word about the actual iodine content of the solution following Jan's comment:

In principle, masses can always be added up (unlike volumes, which should not be added up, due to possible interactions between certain solvents — see here).
Therefore, we can say that the total mass of solutes in one drop is $$\pu{2.5mg}+\pu{3.75 mg} = \pu{6.25 mg}$$ ($\pu{12.5 mg}$ in two drops, per OP question). However, we cannot say that two drops contain $\pu{12.5 mg}$ of iodine, because the solution contains both elemental iodine (I2) and potassium iodide (KI), which are two different constituents. Since KI's role in the solution is to facilitate dissolution of elemental iodine molecules (see description of chemical properties, including equations, here), iodide content is irrelevant and the solution's strength reflects only iodine's content.

In other words, $\pu{12.5 mg}$ is the total quantity of solutes in two drops of Lugol's solution, but this quantity has no practical meaning, since the active ingredient is iodine, and its amount in 2 drops of $5~\%$ Lugol's solution is $\pu{5 mg}$.

  • $\begingroup$ My comment was actually ‘We have three constituents (two solutes, one solvent), so $5~\%$ does not carry any meaning a priori.’ However, thank you for clarifying that a percentage of Lugol’s solution is always an iodide percentage. However, the correct way to calculate solution percentages (not v/v; m/m) would be $\text{mass solute}/\text{mass solution}$ — i.e. we would need to have $m(\ce{I2})/(m(\ce{I2}) + m(\ce{KI}) + m(\ce{H2O}))$. But probably simplicity ruled … $\endgroup$
    – Jan
    Commented Mar 4, 2017 at 0:24

My understanding of the historical formula is that the above amounts are not correct. The 5% Lugol's takes it's name from the approximate percentage of the principle solute which is Iodine. The Potassium Iodide is an important part of the formula. Aside from assisting in the dissolving or the Iodine the Iodide part is an important nutritional factor. Some parts of the body need Iodine and others Iodide. The formula calls for: 10g of Potassium Iodide 5g of Iodine and Have available 100 ml of distilled water

"Dissolve KI in about 20-30 ml of distilled water. Add iodine and heat gently with constant mixing until iodine is dissolved. Dilute to 100 ml with distilled water. Store in amber glass-stoppered bottle in the dark." https://www.fda.gov/Food/FoodScienceResearch/LaboratoryMethods/ucm062245.htm

The confusion in the amounts above is because the Iodide content of 10g of potassium Iodide is about 7.5g

The Iodine content of one metric drop ( =1/20ml) is 2.5mg The Iodide content of one metric drop is 3.35mg The total dose of Iodine plus Iodide in one drop is 6.25mg


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