I would like to know if salt can be dissolved in heavy water ($\ce{D2O}$) as it would be in normal water ($\ce{H2O}$).

If so, what will the maximum salinity be in $\ce{D2O}$? I know the maximum in $\ce{H2O}$ is about 26% of salt by mass.


1 Answer 1


Well, consider a salt as $\ce{AB_s}$ in water you have :

$$\ce{AB}_s\rightleftharpoons \ce{A_{aq}^{q+} + B_{aq}^{q-}}$$

And $\Delta H_{s}=-\ce{U +\Delta H_{aq}}$ where $\ce{U}$ is the lattice energy.

When the salt needs to be solvate, $\ce{O-H}$-bonds of water molecules around this salt need to be rearranged. More, the "$\ce{DO}$-bonds" in $\ce{D_2O}$ are stronger than "$\ce{HO}$-bonds".$^1$ Then it will be harder to rearranged hence it will cost more energy to solvate the salt which will be then less soluble in heavy water than in water.

I found an article in the JACS and as you can see on the image below the maximum solubility of $\ce{KCl}$ in heavy water is a little less than in water (you can make a linear extrapolation) the difference is not quite importante but it looks it exists.

Solubility in water and heavy water too

Look at the second reference $^2$.

$^1$ : https://arxiv.org/ftp/arxiv/papers/0706/0706.1355.pdf, p-5

 Nuclear quantum effects are particularly important in the different properties of light (H2O) and heavy water (D2O) where the more restricted atomic vibrations in D2O reduce the negative effect of its van der Waals repulsive core so increasing its overall hydrogen bond  energy.

$^2$ : For the image, http://pubs.acs.org/doi/abs/10.1021/je60070a011

  • 1
    $\begingroup$ $\ce{O-H}$ bonds in water need to be broken?? If anything, hydrogen bonds are readjusted but no actual bonds are broken in the dissolution of salts … $\endgroup$
    – Jan
    Oct 5, 2016 at 22:23
  • $\begingroup$ I'd say he really mean bond energy especially he does not say bonds in water broken. $\endgroup$ Oct 6, 2016 at 9:24
  • $\begingroup$ @Jan I changed I don't know how to say exactly how to explain but there is an intecration during the solvatation of ions. $\endgroup$
    – ParaH2
    Oct 6, 2016 at 12:00
  • $\begingroup$ This may be far too simplistic but is not that the vibrational zero point energy is smaller in $\ce{D2O}$ compared to water since the vibrational frequency of $\ce{D2O}$ is $\omega =\sqrt {k\mu} $ where ∗k∗ is the force const, same for $\ce{D2O and H2O} $, and $\mu$ the reduced mass, greater for $\ce{D2O}$ thus the bond dissociation energy is greater. This would indicate that to solvate with $\ce{D2O}$ by making new bonds to ions need slightly more energy than for water hence lower solubility at same temperature. $\endgroup$
    – porphyrin
    Oct 7, 2016 at 14:51
  • $\begingroup$ a typo above,, I meant $\omega=\sqrt {k/\mu}$ $\endgroup$
    – porphyrin
    Oct 7, 2016 at 15:30

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