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What is more acidic:

$\ce{D3O+}$ in $\ce{D2O}$

or

$\ce{H3O+}$ in $\ce{H2O}$

and why?

I think it's $\ce{D3O+}$ in $\ce{D2O}$ as I saw somewhere that this property is used in mechanistic studies (the inverse isotope effect), but I need a proper explanation.

EDIT: I found a statement about this in Clayden: Water $\ce{H2O}$ is a better solvating agent for $\ce{H_3O^+}$ than $\ce{D2O}$ is for $\ce{D_3O^+}$ because O–H bonds are longer than O–D bonds hence $\ce{D_3O^+}$ in $\ce{D2O}$ is stronger than $\ce{H_3O^+}$ in $\ce{H2O}$. That is the origin of the inverse solvent isotope effect.

This is in contrast with all the answers given yet, so I am pretty confused.

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  • $\begingroup$ related chemistry.stackexchange.com/questions/31587/… $\endgroup$ – Mithoron Nov 25 '15 at 18:43
  • $\begingroup$ en.wikipedia.org/wiki/… here you can see that pH of $\ce{D2O}$ is 7.43 which is more than that of water...therefore light water is more acidic. but I dont have the explaination. $\endgroup$ – manshu Nov 25 '15 at 19:01
  • $\begingroup$ the pH of heavy water should be more than that of normal H2O bcoz it is difficult to break the DO bond $\endgroup$ – manshu Nov 25 '15 at 19:10
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    $\begingroup$ But the pKw of D2O isn't 14 right? $\endgroup$ – RBW Nov 25 '15 at 19:20
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For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.

The entire point of the above reference is that

$\ce{H3O+ + H2O <=> H2O + H3O+}$

(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.

$\ce{D3O+ + D2O <=> D2O + D3O+}$

would also correspond to an equilibrium constant of 1

So when Clayden and the OP write

$\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$

it is wrong for the above reason.

Two genuine thermodynamic equilibriums are

$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$

Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).

But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$. $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.

As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.

Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.

Also the equilibrium constant for

$\ce{H2O + H2DO+ <=> HDO + H3O+}$

can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751


Explanation of Normal/Inverse Solvent Isotope Effect

For a kinetic normal/inverse solvent isotope effect there will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.

Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.

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    $\begingroup$ But "why" is this the case? $\endgroup$ – ron Nov 25 '15 at 20:36
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    $\begingroup$ @ron I don't think anyone knows the complete reasons. In “Quantum Differences between Heavy and Light Water.” Physical Review Letters 101 065502 (2008) it is said that D2O is stabilized by more hydrogen bonds per molecule (3.76 vs. 3.62) phys.org/news/2008-08-scientists-quantum-differences-heavy.html $\endgroup$ – DavePhD Nov 25 '15 at 22:02
  • $\begingroup$ So what's then the explanation of the inverse isotope effect? $\endgroup$ – RBW Nov 27 '15 at 18:54
  • $\begingroup$ @Marko section 8.1.6 here: books.google.com/… is better than anything I can think of at the moment $\endgroup$ – DavePhD Nov 27 '15 at 19:30
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This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$

\begin{aligned} \ce{H2O + H2O &~<=> H3O^+ + OH^{-}&&(I)}\\ \ce{D2O + D2O &~<=> D3O^+ + OD^{-}&&(II)}\\ \end{aligned}

The reason why this is the case is the interesting question.

Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.

  • On the reactant side, $\ce{D2O}$ is more stable than $\ce{H2O}$ since an $\ce{O-D}$ bond will have a lower zero-point energy than the corresponding $\ce{O-H}$ bond. Hence the deuterated molecule will have a lower overall energy compared to its protonated analogue.
  • On the product side, the deuterated ion must again be more stable than the protonated ion for the same zero-point energy reasons. However, since the bonds in the $\ce{X3O^+}$ ions are weaker (longer) than the bonds in the neutral reactants, the zero-point energy differences would be compressed or lessened.

The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.

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  • $\begingroup$ Could you elaborate a little this conclusion: Generally, acidic compounds (phenols, carboxylic acids, etc.) are more dissociated in D2O than H2O (ref). This would suggest that the equilibrium constant in eqn (I) would be larger than the equilibrium constant in eqn (II) $\endgroup$ – RBW Nov 25 '15 at 21:20
  • $\begingroup$ @ron are you sure the bonds of the ions are longer? In liquid, H2O has O-H bond length of 1.01 Angstroms by diffraction and D2O has O-D = 0.98 Angstroms. For hydronium I see 1.002 A calculated in liquid www1.lsbu.ac.uk/water/hydrogen_ions.html and smaller lengths for hydroxide www1.lsbu.ac.uk/water/ionisoh.html but not clear if this is in liquid. $\endgroup$ – DavePhD Nov 25 '15 at 23:56
  • $\begingroup$ No, there's really not much that I'm sure about. By "ion" I meant the $\ce{X3O^+}$ species, not the hydroxide, and would expect the weaker bonds in those $\ce{X3O^+}$ ions to also be longer. I've edited the answer to clarify that point. $\endgroup$ – ron Nov 26 '15 at 0:01
  • $\begingroup$ Therefore D3O+ in D2O is stronger, right? $\endgroup$ – RBW Nov 26 '15 at 13:48
  • $\begingroup$ No, equilibrium II lies further to the left than equilibrium I, therefore the $\ce{D2O}$ case is less acidic. $\endgroup$ – ron Nov 26 '15 at 14:26
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When comparing $\ce{H2O}$ and $\ce{D2O}$, I can't help but think that the biggest different between the two molecules is their vibrational properties. Specifically, $\ce{H2O}$ is going to vibrate at a frequency which is $\sqrt2$ times larger than that of $\ce{D2O}$ based on$$\nu=\frac{1}{2\pi}\sqrt\frac{{\kappa}}{{\mu}}$$ because they will have virtually the same bond force constant , but the reduced mass of a $\ce{DO}$ bond is (basically) twice that of $\ce{H2O}$.

Thus, I would speculate that an explanation for why $\ce{H2O}$ might be more easily deprotonated lies in the fact that it vibrates at a substantially higher frequency and so it is instantaneously in higher energy states and thus closer to overcoming the energy barrier intrinsic to breaking a bond.

This of course says nothing about the relative stabilities of $\ce{H3O+}$ and $\ce{D3O+}$, but ron deals with that issue in his answer quite well and his analysis corresponds with the claim of my answer too.

It would be interesting to know if tritium is less acidic than deuterium (assuming it is) at a proportion which is smaller than that same proportion between deuterium and protium because that idea would align with the vibration explanation I believe. I can't find that data anywhere though.


This website has some very accessible information on liquid water vibrations if anyone is interested.

Also, feel free to disagree. I'm not completely sold on this suggestion, but it seems like a valid explanation.

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  • $\begingroup$ a DO bond has a "reduced mass"? $\endgroup$ – Dissenter Nov 26 '15 at 1:23
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    $\begingroup$ Ya. That's what $\mu$ represents. It's a way of reducing a two body problem into a one body problem which is commonly used for vibrating systems. Essentially you're looking at the vibration of the center of mass rather than the vibration of the two atoms. So, all bonds have a reduced mass not just OD. Sorry if that was unclear. $\endgroup$ – jheindel Nov 26 '15 at 1:47
  • $\begingroup$ Nice, I think we are describing the same thing on the "reactant" side. The zero-point energy I refer to is what you are describing using the reduced mass and vibrational energy, see here for example $\endgroup$ – ron Nov 26 '15 at 2:28
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Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.

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