Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:
The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:
omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and
taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.
With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$. For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.
However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:
"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.
The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.
Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:
The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:
omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and
taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.
With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$. For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.
However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:
"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.
The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.
So it sounds like the origin of the -1.76 value arises from a simplified definition of pKa often used in introductory courses. In any case, what implications does this have, if the pKa is really 1.0 instead of -1.76? Does the pKa of protonated methanol differ significantly?More
@Dissenter "what implications does this have, if the pKa is really 1.0 instead of -1.76" Nobody is saying 1.0, the only values mentioned are 0.0, -0.7 and -1.74. But I dont think there is any consequences because, as the article explains, the only real equilibrium with H2O and H3O+ is Kw. I cant think of any situation where Ive ever used a value for Ka of H3O+ in water.More
@Dissenter "I found that the pKa of protonated methanol to be -2.2 Would H3O+ really be that much different from protonated methanol?" In DMSO, water pKa is 2.4 units higher than the methanol pKa More
The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.
There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.
The closest IUPAC defined term is the standard equilibrium constant, which can be denoted as $K^\circ$ or just $K$.
Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:
$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$
Where $a$ is activity.
Substituting that the acid is $\ce{H3O+}$:
$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$
and of course $-\log(1) = 0$.
The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:
With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$. For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.
However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:
New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water, which insists that
$$\ce{H2O + H3O+ <=> H3O+ + H2O}$$
"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.
The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.
The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.
There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.
The closest IUPAC defined term is the standard equilibrium constant, which can be denoted as $K^\circ$ or just $K$.
Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as:
$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$
Where $a$ is activity.
Substituting that the acid is $\ce{H3O+}$:
$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$
and of course $-\log(1) = 0$.
The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:
With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$. For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.
However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:
New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water, which insists that
$$\ce{H2O + H3O+ <=> H3O+ + H2O}$$
"does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.
The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.
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I think there is a misunderstanding between Keq and Ka:
for $\ce{H3O+ + H2O <=> H3O+ + H2O}$:
$$K_{eq} = \frac{[\ce{H3O+}]*[\ce{H2O}]}{\ce{[H3O+]*[H2O]}} = 1$$
however $K_\mathrm{a} = K_{eq}*[\ce{H2O}] = \ce{[H2O]}$
therefore $K_\mathrm{a}(\ce{H3O+}) = [\ce{H2O}]$
p$K_\mathrm{a}(\ce{H3O+}) = -1.74$
I think there is a misunderstanding between Keq and Ka:
for $\ce{H3O+ + H2O <=> H3O+ + H2O}$:
$$K_{eq} = \frac{[\ce{H3O+}]*[\ce{H2O}]}{\ce{[H3O+]*[H2O]}} = 1$$
however $K_\mathrm{a} = K_{eq}*[\ce{H2O}] = \ce{[H2O]}$
therefore $K_\mathrm{a}(\ce{H3O+}) = [\ce{H2O}]$
p$K_\mathrm{a}(\ce{H3O+}) = -1.74$
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