Deuteronium deuteroxide. The why of pD 7.435.

Earlier, I constructed a possible model of hydronium hydroxide, or H3O+.OH– One way of assessing the quality of the model is to calculate the free energy difference between it and two normal water molecules and compare the result to the measured difference. Here I apply a further test of the model using isotopes.

Pure water has pH 7, which means equal concentrations for both [H3O+] and  [OH] of 10-7M. Converting this to a free energy one gets ΔG298 19.088 kcal/mol. Now the pD of pure deuterium oxide is reported as 7.435, equivalent to ΔG298 20.274, an isotope effect on the free energy of ΔΔG298 =1.186 kcal/mol. How does the theoretical model (ωB97XD/Def2-TZVPPD/SCRF=water) previously reported[1],[2] do? The value obtained is 1.215,[3] an apparent error of only 0.029 kcal/mol. I am quite pleased with the close correspondence; at least the model is capable of reporting good isotope effects on the ionisation equilibrium of pure water!

Finally, with some confidence assured, one might apply this to tritonium tritoxide. Tritiated water is so radioactive it would boil in an instant, probably well before its pT could be measured. ΔΔG298 is calculated as 1.798 kcal/mol. Will this estimate ever be challenged by experiment?


‡ It is assumed no isotope effect acts on the dielectric constant of water and hence the continuum model used here to model it. In fact the isotope effect on this property is modest; ε298 = 77.94, compared with 78.36 for normal water.[4]

References

  1. Henry S Rzepa., "H 22 O 11", 2016. http://dx.doi.org/10.14469/ch/191999
  2. Henry S Rzepa., "H 22 O 11", 2016. http://dx.doi.org/10.14469/ch/191998
  3. Henry Rzepa., "Deuteronium deuteroxide; free energy differences.", 2016. http://dx.doi.org/10.14469/hpc/407
  4. C. Malmberg, "Dielectric constant of deuterium oxide", Journal of Research of the National Bureau of Standards, vol. 60, pp. 609, 1958. http://dx.doi.org/10.6028/jres.060.060

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4 Responses to “
Deuteronium deuteroxide. The why of pD 7.435.

  1. look@s says:

    The pKw values of H2O, D2O and T2O are given as 14.000, 14.869 and 15.215 in "Lehrbuch der Anorganischen Chemie" by Holleman/Wiberg.  Not sure of the source, but here is an article where they measure the triple point of T2O after preparing it from tritium gas over copper oxide:  http://pubs.acs.org/doi/pdf/10.1021/ja01143a070 The sample is first frozen and then warmed ("The sample warmed up partly by the energy release associated with the radioactive decay of the tritium."), but the heat evolution does not seem to be particularly strong.

  2. Henry Rzepa says:

    Yes, 2*7.435 = 14.87, which is indeed the value you give. The pKw of T2O comes out as a free energy 20.744 kcal/mol, an isotope effect of 1.656 kcal/mol (vs 1.798 calc).  I presume the values you quote are at 298K? Are they corrected for small residual amounts of H, which might reduce the value slightly?

    The assumed fact was always that pure T2O would boil, but that is clearly not true! 

  3. look@s says:

    Temperature is indeed 298 K. Not sure about errors. The only article I found about measuring the value for T2O is from 1969 http://dx.doi.org/10.1063/1.1672273 and gives K = 6.1 E-16, which corresponds exactly to the 15.215 given above.

  4. Henry Rzepa says:

    H2O  is predicted to give about 20 times as much H+ as T2O gives T+. Thus even a few % of H2O present in T2O would increase the apparent ionization. K = 6.1 E-16 therefore might be taken as an upper limit to the ionisation.

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