What is the pKa of the O-17 hydronium ion?

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](https://chemistry.stackexchange.com/q/9847).


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:


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


