I understand that tap water contains minute amounts of deuterium oxide (heavy water). Is there a simple way to separate, concentrate the $\ce{D2O}$?

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    $\begingroup$ No, there is no simple way. Wikipedia or a Google search (heavy water separation process) will lead you to some references. The main point is that the natural abundance of D is only ~110 parts per million, so you need very large scale industrial processes to separate the D from H. Also, since the abundance of D is so low, there is almost no $D_{2}O$ in nature - it will almost all be $DHO$. $\endgroup$
    – Jon Custer
    Nov 17, 2014 at 18:05
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    $\begingroup$ Agree with Custer. You need a super-elaborate distillation process, since HOD apparently boils at 101.4. $\endgroup$
    – Lighthart
    Nov 17, 2014 at 20:00
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    $\begingroup$ The biggest production plant used the Girdler Sulfide process, which exploited equilibria between deuterated hydrogen sulfide and deuterated water. That's probably the way to go (but not in your house. Hydrogen sulfide is nasty AND toxic). $\endgroup$
    – AlaskaRon
    Nov 6, 2015 at 9:31

3 Answers 3


It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)

Electrolysis of water strongly favors "H" being converted to hydrogen gas rather than "D", by a factor of about 8 to 1 (depending upon the electrodes).

Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$ Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method. The article says the process is suitable for "those who have occasion to prepare or concentrate heavy water for research, demonstration or advanced student laboratory use".

In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains. This increased the original concentration of deuterium by a factor of 12.

By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage.

Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level. They used three stages. Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.


Basically you freeze water partially. DDW has a higher freezing temperature. Stick a bowl of water in freezer. When the top freezes over wait another 15 minutes. Take out of freezer and poke a hole in ice. The water that comes out is around 50 ppm. If you do this a few times you can get it down even further BUT: this method does not remove all impurities from ddw. If you start with filtered water first (like a brita fliter on your faucet) you'd be much better off.

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    $\begingroup$ What’s DDW?$%..$ $\endgroup$
    – Jan
    Aug 29, 2017 at 4:10
  • $\begingroup$ Haha, I'm sure this would work in principle, but you need to optimise the process quite a bit before you get, say, 50% D2O that way in a reasonable number of steps. And btw. you remove NO impurities this way. They always accumulate in the liquid phase. $\endgroup$
    – Karl
    Nov 25, 2022 at 18:37

Because, Deuterium is twice as Heavy as Protium (the more abundant form of Hydrogen) it gives the molecule of water a greater moment of inertia and slows down its average rate of rotation. Consequently it does not bond to an growing ice crystal quite as rapidly as the lighter molecules of water. So if you bring water to a boil (to break up polymerization and increase the number of single water molecules) and then cool it rapidly to the point of fusion so there is less time for them to participate in polymer groups the ice will have a greater percentage of light molecules than normal water and the water that is not frozen will have a slightly higher percentage of deuterium. Instead of 110 ppm you might get it to 112 parts per million. So the water that didn't freeze is brought to a boil again and then rapidly cooled to the point of fusion removing even more of the lighter molecules. A very slow process with the only difficulty being the rapid cooling needed to avoid excessive polymerization. It requires far simpler apparatus than distillation or electrolysis as given in the first answer, but is much less efficient and the boiling phase would eventually use more energy. However, the question gave simmplicity as a criteria and this boil and freeze process is simple.

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    $\begingroup$ Obscure description with rather misused technical terms. $\endgroup$
    – Greg
    Nov 6, 2015 at 4:49
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ Nov 6, 2015 at 6:03
  • $\begingroup$ I think you mean freezing point not point of fusion. Point of fusion is something quite different. $\endgroup$
    – bon
    Nov 6, 2015 at 14:25
  • $\begingroup$ @bon how is it different? merriam-webster.com/dictionary/fusion%20point $\endgroup$
    – DavePhD
    Mar 13, 2017 at 14:03

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