# How much D2O is in Vienna Standard Mean Ocean Water?

I'm curious how much $$\ce{D2O}$$ (heavy water) is in Vienna Standard Mean Ocean Water. I am interested in $$\ce{D2O}$$ molecules and not $$\ce{HDO}$$/$$\ce{D+}$$/ $$\ce{DH2O+}$$/$$\ce{D2HO+}$$/$$\ce{D3O+}$$ or any other deuterium species or ions. Vienna Standard Mean Ocean Water is pure water extracted from some ocean with a specific isotopic composition. Sweet water has slightly different isotopic composition.

A suitable unit for the result could be mol/kg or a mol fraction.

I'm not sure if temperature or pressure make a difference. I suggest $$\pu{20 ^\circ C}$$ (or any other temperature, please specify, if possible) and $$\pu{101325 Pa}$$, but other conditions are ok, too. Liquid water makes most sense, I guess.

Allright, I took some effort as suggested and used my brain ...

2H/1H = 155.76 ± 0.1 ppm (a ratio of 1 part per approximately 6420 parts)1

So if an HDO autoprotolyzed ...

HDO --> D+ + OH- or HDO --> H+ + OD-

and then D+ or OD- reformed water with OH- or H+, then it would have a 1 in 6420 chance of forming D2O, and so D2O would be 1 in 6420^2=41,216,400 parts.

Sounds good?

• Not too much, $\approx$ 1:36000000 . Oct 9, 2020 at 10:56
• This question makes no sense. Please elaborate, and show your own effort.
– Karl
Oct 9, 2020 at 11:14
• The question makes sense, butI agree with the lack of effort while the asnwer has the obvious way to obtain. Oct 9, 2020 at 12:25