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What is the $\ce{K_{a}}$ value for $\ce{H^+}$. I understand that the hydrogen proton doesn't stick around in solution by itself for very long. Nonetheless, does it have a $\ce{K_{a}}$ value? Can it have a $\ce{K_{a}}$? I think it can have one but quantifying it might be hard. Its $\ce{K_{a}}$ value I think would correspond to this reaction:

$\ce{H^+ + H_2O ->H_3^+O}$

However this doesn't look like a conventional acid/base reaction; we don't have the conjugate base of $\ce{H^+}$ here, unless the definition of a conjugate base isn't what I think it is (acid missing one proton).

So my questions are:

1) Does the above mentioned $\ce{K_{a}}$ value exist.

2) Is my definition of a conjugate base too narrow?

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up vote 2 down vote accepted

The equillibrium between a proton plus water molecule and a hydronium ion is better considered using the concept of proton affinity.

\begin{aligned} \ce{H^+(g) &-> H^+(aq)} \\ \Delta E &= −1530~\mathrm{kJ/mol} \end{aligned}

You could roughly convert to an equillibrium constant using

$$\Delta G = -RT \ln(K)$$

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Would $K$ be defined as $$K=\frac{a(\ce{H3+O})}{a(\ce{H+})\cdot{}a(\ce{H2O})}?$$ And then consequently converted into concentrations via $$c(\ce{H+})=\frac{c^\circ}{\gamma}\cdot{}a(\ce{H+})?$$ Ultimately leading to $$K_a\approx\frac{a(\ce{H3+O})}{a(\ce{H+})}?$$ Might be a bit of a stretch? – Martin - マーチン May 30 '14 at 4:02

I don't think $\ce{K_{a}}$ has meaning in that context. $\ce{K_{a}}$ is defined as the equilibrium constant for the following dissociation reaction $$\ce{HA{<=>}H^+ + A^-}$$

$\ce{H^+}$ can't dissociate; that is, there is no "$\ce{A}$" in the case your considering.

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Good point; there is no acid dissociation! Good way of looking at the problem! – Dissenter May 29 '14 at 20:47

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