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Context

The answer to How to calculate Ka for hydronium and Kb for hydroxide?, water was regarded as solvent once and base (or equivalently acid) once while citing Reference 1 (1986) and made no comments on the dissociation constants of $\ce{H3O+}$ and $\ce{HO-}$ in water.

$$ K_{\text{a,} \ce{H2O}} = K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}} \approx \dfrac{10^{-14}}{\ce{[H2O]}} = ; \text{room temperature} $$

The question was marked duplicate to What is the pKa of the hydronium, or oxonium, ion (H3O+)?, and the answer disregarded auto-protonation of water as a thermodynamic process while citing Reference 2 (1998). However, since the reference did not consider it, the answer also did not contain any information on the dissociation constant of $\ce{HO-}$.

I don't think these questions are duplicate since these have been answered differently, and neither provide dissociation constants for both $\ce{H3O+}$ and $\ce{HO-}$. Moreover, neither question or answers have considered the relation:

$$ K_\text{a}K_\text{b} = K_\text{w} $$

I think there is room for further discussion here.

Question

Are there any more recent developments on this topic? Has a definitive answer been provided somewhere?

References

  1. Starkey, R., Norman, J., Hintze, M. (1986). Ronald Starkey, Jack Norman, and Mark Hintze. J. Chem. Educ., 63(6), 473. 10.1021/ed063p473
  2. Burgot, J. (1998) New point of view on the meaning and on the values of Ka○(H3O+, H2O) and Kb○(H2O, OH–) pairs in water. Analyst,123, 409-410. 10.1039/A705491B
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Housecroft and Sharpe$^\text{3}$

Formulation of $K_\text{a}$ and $K_\text{b}$:$^\text{3}$

Ka and Kb formula

This seems to side towards Reference 1 (1986) and not Reference 2 (1998):

For general reactions $\ce{HA + H2O -> H3O+ + A-}$ and $\ce{B- + H2O -> HB + HO-}$, $K_\text{a}$ and $K_\text{b}$ are given by:

$$ \dfrac{\ce{[H3O+][A-]}}{\ce{[HA]}} \equiv \dfrac{[\text{Conjugate Acid}][\text{Conjugate Base}]}{[\text{Acid}]} \tag{1} $$

and

$$ \dfrac{\ce{[HB][HO-]}}{\ce{[B-]}}\equiv \dfrac{[\text{Conjugate Acid}][\text{Conjugate Base}]}{[\text{Base}]} \text{,} \tag{2} $$

respectively.

Self-Ionization of Water

Consider the following:

$$ \ce{\underset{\text{acid}}{H2O} + \underset{\text{base}}{H2O} -> \underset{\text{conjugate acid}}{H3O+} + \underset{\text{conjugate base}}{HO-}} $$

$K_{\text{a,} \ce{H2O}}$, from Equation (1), is given by:

$$ K_{\text{a,} \ce{H2O}} = \dfrac{\ce{[H3O+][HO-]}}{\ce{[H2O]}} = \dfrac{K_\text{w}}{\ce{[H2O]}} $$

$K_{\text{b,} \ce{H2O}}$, from Equation (2) is given by:

$$ K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}} $$

An Example

Consider this example:

Calculation of Kb of ammonia from Kw and Ka of ammonium

Calculation for $\ce{H3O+}$

If we consider our acid as $\ce{H3O+}$:

$$ \ce{H3O+ + H2O -> H3O+ + H2O}\implies K_{\text{a,}\ce{H3O+}} = \dfrac{\ce{[H3O+][H2O]}}{\ce{[H3O+]}} = \ce{[H2O]}\\ $$

This is in agreement with $K_\text{a}K_\text{b} = K_\text{w}$

$$ \implies K_{\text{a,}\ce{H3O+}}K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}}\ce{[H2O]} = K_\text{w} $$

Calculation for $\ce{HO-}$

If we consider our base as $\ce{HO-}$:

$$ \ce{HO- + H2O -> H2O + HO-}\implies K_{\text{b,}\ce{HO-}} = \dfrac{\ce{[HO-][H2O]}}{\ce{[HO-]}} = \ce{[H2O]} $$

This is in agreement with $K_\text{a}K_\text{b} = K_\text{w}$

$$ \implies K_{\text{a,}\ce{H2O}}K_{\text{b,} \ce{HO-}} = \dfrac{K_\text{w}}{\ce{[H2O]}}\ce{[H2O]} = K_\text{w} $$

Remarks

I am siding with Housecroft and Sharpe and Reference 1 (1986) and not Reference 2 (1998) because of the agreement with $K_\text{a}K_\text{b} = K_\text{w}$. However, this would also mean:

$$ \boxed{\mathrm{p}K_{\text{a,} \ce{H3O+}} = \mathrm{p}K_{\text{b,} \ce{HO-}} \approx 3.16 \times 10^{-56}} $$

as stated in Reference 1.

References

  1. Starkey, R., Norman, J., Hintze, M. (1986). Ronald Starkey, Jack Norman, and Mark Hintze. J. Chem. Educ., 63(6), 473. 10.1021/ed063p473
  2. Burgot, J. (1998) New point of view on the meaning and on the values of Ka○(H3O+, H2O) and Kb○(H2O, OH–) pairs in water. Analyst,123, 409-410. 10.1039/A705491B
  3. Housecroft C. E., Sharpe A.G. (2008). Inorganic Chemistry, 3$^\text{rd}$ ed. Pearson Education Ltd.
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  • 1
    $\begingroup$ Ananta, You appear to be a very keen student of science which is good. This is not for criticism but what I have learned over time. Science, chemists, textbook authors do not know everything. However, sometimes these debates become debates of questions like, "How many angels can land on a pinhead"? Water is one of the miracles of creation. Its bulk structure is still full of active research in physical , and there is perhaps nothing such as $\ce{H3O+}$ in water. Chemists do have a tendency to talk about a lot of stuff without having any decent knowledge about it. $\endgroup$
    – ACR
    Commented Jun 8, 2023 at 3:16
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    $\begingroup$ @AChem I never knew about non-existence of $\ce{H3O+}$, this would be a very interesting topic to discover. Could you provide some citation that could guide me to it? $\endgroup$
    – ananta
    Commented Jun 8, 2023 at 3:18
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    $\begingroup$ It is certainly an interesting topic but out of my area of expertise but I have attended several advanced talks on the structure of water which perplexes best of the best. What I am trying to teach is to be skeptical. By that I mean don't take any "answer" as final. I meant that $\ce{H3O+}$ does not exist independently, it is surrounded by many many water molecules, in a dynamic structure. See solvation of hydronium here: en.wikipedia.org/wiki/Hydronium. One may discuss hydronium in non-aqueous systems, then it makes more sense. See pubs.acs.org/doi/abs/10.1021/ja00508a040 $\endgroup$
    – ACR
    Commented Jun 8, 2023 at 3:26
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    $\begingroup$ @AChem thank you. Although I have provided an answer, i don't consider it as the answer. As I stated in the question, I think there is room for further discussion here. I am as skeptical as a sparrow going to pick a grain next to a snake's lair on this one. $\endgroup$
    – ananta
    Commented Jun 8, 2023 at 3:34
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    $\begingroup$ Many chemists consider the hydratation of $\ce{H+}$ goes much further than $\ce{H3O+}$, so that $\ce{H3O+}$ must be replaced by $\ce{H5O2^+}$ (with a proton between two $\ce{H2O}$ molecules), or even more complicated structures like $\ce{H13O6^+}$ (with one proton between six $\ce{H2O}$ molecules). Serious arguments may be given in favor of these complicated structures. But for the ordinary analyst, they are not fundamental. For titration purposes, it is sufficient to know that the acidic proton is hydrolyzed, whatever the number of water molecules attached to it. $\endgroup$
    – Maurice
    Commented Jun 8, 2023 at 9:55

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