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The familiar two-step $\mathrm{E2}$ mechanism of the dehydration reaction of ethanol using concentrated sulfuric acid is given below (forgive me for not using diagrams):

  1. $\ce{H-CH2-CH2-OH + H2SO4 -> H-CH2-CH2-OH2+ + HSO4-}$
  2. $\ce{HSO4- + H-CH2-CH2-OH2+ -> H2SO4 + CH2=CH2 + OH2}$

I have certain doubts about the viability of the second step, as $\ce{HSO4-}$ is a weak base.

Maybe $\ce{H2O}$ is used as the base in the second reaction instead? If so, is the catalyst regenerated at all?

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What actually happens, as always, depends on the exact conditions and concentrations which you did not add to your question. However, you do state whether the catalyst is regenerated; as such, I assume you are talking about the following reaction:

$$\ce{C2H5OH ->[\ce{[H2SO4]}] C2H4 + H2O}\tag{1}$$

I.e. a dehydration of ethanol using catalytic sulfuric acid.

The first thing we have to consider is the $\mathrm{p}K_\mathrm{a}$ values of sulfuric acid and ethanol (and water; which will be present anyway unless you accidentally found dry sulfuric acid). Using a very quick ballpark, we can assume: $$\mathrm{p}K_\mathrm{a}(\ce{H2O}) \approx \mathrm{p}K_\mathrm{a} (\ce{C2H5OH}) > \mathrm{p}K_\mathrm{a}(\ce{H2SO4})\tag{2}$$

Since sulfuric acid is catalytic, we can assume that the first (complete; non-equilibrious) step will be:

$$\ce{C2H5OH + H2SO4 -> C2H5OH2+ + HSO4-}\tag{3}$$

$\mathrm{p}K_\mathrm{a,2} (\ce{H2SO4}) \approx 2$, which is only two units less than water. Therefore, — under the assumption that we added $10~\mathrm{mol}\text{-}\%\ \ce{H2SO4}$ — we might assume that some additional molecules of ethanol (and/or water) are protonated under equilibrium conditions:

$$\ce{C2H5OH + HSO4- <<=> C2H5OH2+ + SO4^2-}\tag{4}$$

Thus, part of the sulfuric acid will be present as sulfate, the majority as hydrogen sulfate. Sulfate is a stronger base (albeit still weak) and thus the final mechanistic step is probably better written using sulfate rather than hydrogen sulfate:

$$\ce{H-C2H4-OH2+ + SO4^2- -> HSO4- + H2O + C2H4 ^}\tag{5}$$

Of all the bases present at equilibrium, sulfate is the strongest as its conjugate acid’s $\mathrm{p}K_\mathrm{a} > 0$. The general order of basicity of all compounds in solution is:

$$\ce{SO4^2-} > \ce{H2O} \approx \ce{C2H5OH} > \ce{HSO4-}\tag{6}$$

Finally, note that the reaction does not have to proceed via $\mathrm{E2}$ — an $\mathrm{E1}$ mechanism is almost equally viable.

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