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At low concentrations of a conjugate acid/base pair in aqueous solutions, the Henderson-Hasselbalch equation gives a good approximation of the ratio of acid and base:

$$ \frac{[\ce{A-}]}{[\ce{AH}]} = 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})\tag{1}}$$

If I add a low concentration of $\ce{[^17O]}$ water to an aqueous solution with pH of zero, I would think the heavy water behaves similarly to the solvent. In fact, we don't even have to add heavy water because at natural abundance, 0.038% of oxygen atoms are isotope 17.

I am wondering what the ratio of hydronium to water is for the heavier isotope. For the solvent, it is about 1:55. If it is the same for the heavier isotope, the reaction

$$\ce{[^17O]H2O + H+ <=> [^17O]H3O+}\tag{2}$$

or

$$\ce{[^17O]H2O(aq) + H3O+(aq) <=> [^17O]H3O+(aq) + H2O(l)}\tag{3}$$

would have a $\mathrm{p}K_\mathrm{a}$ of -1.74. However, the $\mathrm{p}K_\mathrm{a}$ of water is sometimes quoted as 0.

So is the $\mathrm{p}K_\mathrm{a}$ of $\ce{[^17O]H2O}$ different than that of $\ce{[^16O]H2O}$, or do I have to use two different values, depending on whether the solvent is something else or part of the acid/base reaction in question. Surely, $\ce{[^17O]H2O}$ and $\ce{[^16O]H2O}$ have similar $\mathrm{p}K_\mathrm{a}$ values in DMSO, where the equilibrium reaction shown above (3) would be written slightly differently and would have an equilibrium constant very close to one:

$$\ce{[^17O]H2O(DMSO) + H3O+(DMSO) <=> [^17O]H3O+(DMSO) + H2O(DMSO)}\tag{4}$$

Karsten
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  • I would use -1.74 for O-17 water and 0 for O-16 water if O-16 is the bulk solvent, because it seems like those are the values that would give the right results for any relevant calculation, but I'm not confident that's the correct general answer – Andrew Apr 28 '22 at 19:21
  • The difference between 0 and -1.74 has nothing to do with isotopes. It has to do with what we define as the unit quantities to put into the equilibrium constant. If a unit corresponds to the pure solvent fir water, then the $pK_a$ is zero. If the unit is 1 molar concentration for water as for other species, then pure water is about 55 units and from the logarithm of that we end with -1.74. It's a confusing convention, but at lease we can say that either way using oxygen-17 should have hardly any effect. – Oscar Lanzi Apr 28 '22 at 19:25
  • @OscarLanzi I agree, but this specific example shows that there are cases where the value of -1.74 does make sense. If you switched the concentrations of heavy and light water (i.e. traces of O-16 water in bulk O-17 water), you would get zero for O-17 and -1.74 for O-16 with the commonly used standard state definitions (or say that in this case, the $\mathrm{p}K_\mathrm{a}$ of O-16 water is not defined, but the Kw is). – Karsten Apr 30 '22 at 01:09

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