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This question already has an answer here:

I am wondering how to calculate the $K_a$ for hydronium, $H_3O^+$, and the $K_b$ for hydroxide. Because hydronium undergoes the equilibrium

$\ce{H_3O^+(aq) + H_2O (l) <=> H_2O (l) + H_3O^+ (aq)}$ ,

isn't the constant exactly 1? (And the same thing for hydroxide?) However, Wikipedia gives the $pK_a$ for hydronium to be -1.7. Does it have something to do with the "concentration" of water, which is 55.6 M, and upon taking the negative natural log, is -1.74?

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marked as duplicate by orthocresol, Todd Minehardt, Jannis Andreska, Klaus-Dieter Warzecha, bon Mar 5 '16 at 20:00

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To an extent, you are right; the calculation of the $K_\mathrm{a}(\ce{H2O})$ does include $[\ce{H2O}]$.

Consider the following equation: $$\ce{H2O(l) +H2O(l) ->H3O+(aq) + OH- (aq)}$$

The equilibrium for this equation is:

$$K_\mathrm{a}(\ce{H2O})=\frac{[\ce{H3O+}]\cdot [\ce{OH-}]}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Since $K_\mathrm{w}=[\ce{H3O+}]\cdot [\ce{OH-}]=1.01 \times 10^{-14}$ at $25^\circ \mathrm{C}^{[1]}$, we can make such a substitution in our equation:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Now here you will have to bear with me. In a usual $K_\mathrm{a}$ calculation, $[\ce{H2O}]$ is disregarded, as it remains mostly constant ($55.5084\ \mathrm{M}$), except when working with extremely concentrated solutions. In this equation, there are two $[\ce{H2O}]$ terms, and only one of them is disregarded (as a solvent). The other one can be considered concentration of the acid. Therefore:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]}=\frac{1.01 \times 10^{-14}}{55.5084}=1.819544 \times 10^{-16}$$

Thus we have calculated the $K_\mathrm{a}$ for water, or the $K_\mathrm{b}$ for $\ce{H3O+}$. The relation between these is:

$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$

So:

$$K_\mathrm{a}(\ce{H3O+})=\frac{1.01 \times 10^{-14}}{1.819544 \times 10^{-16}}=55.5084$$

and: $$-\log(K_\mathrm{a}(\ce{H3O+}))=\mathrm{p}K_\mathrm{a}(\ce{H3O+})=-\log(55.5084)=-1.7404$$

The calculation for the $K_\mathrm{b}(\ce{OH-})$ is exactly the same for the calculation of $K_\mathrm{a}(\ce{H3O+})$ (pretty much just replace all the $\ce{H3O+}$ with $\ce{OH-}$) and yields the same result.


[1] Starkey, R.; Norman, J.; Hintze, M. Who Knows the Ka Values of Water and the Hydronium Ion? J. Chem. Educ. Journal of Chemical Education. 1986, 63, 473.

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Including water on both side of the equation is a bit of a mistake here.

For hydronium:

$$\ce{K_a = \frac{[H^+][H_2O]}{[H_3O^+]}}$$

and the source of the proton in question is not likely water (normally a stronger mineral acid).

If you put the water on both sides of the equation you are describing another process, which would be proton transfer equilibrium. In that case, you are correct the constant would be 1, but all that means is that any water is as good as any other water at moving the proton around.

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