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?

1 https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water

  • $\begingroup$ Not too much, $\approx$ 1:36000000 . $\endgroup$
    – peterh
    Oct 9, 2020 at 10:56
  • 2
    $\begingroup$ This question makes no sense. Please elaborate, and show your own effort. $\endgroup$
    – Karl
    Oct 9, 2020 at 11:14
  • 3
    $\begingroup$ The question makes sense, butI agree with the lack of effort while the asnwer has the obvious way to obtain. $\endgroup$
    – Poutnik
    Oct 9, 2020 at 12:25


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