E2 step in dehydration of primary alcohol using concentrated sulfuric acid
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:

$$\ce{C2H5OH -&gt;[\ce{[H2SO4]}] C2H4 + H2O}\tag{1}$$

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: $$\mathrm{p}K_\mathrm{a}(\ce{H2O}) \approx \mathrm{p}K_\mathrm{a} (\ce{C2H5OH}) &gt; \mathrm{p}K_\mathrm{a}(\ce{H2SO4})\tag{2}$$

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

$$\ce{C2H5OH + H2SO4 -&gt; C2H5OH2+ + HSO4-}\tag{3}$$

$\mathrm{p}K_\mathrm{a,2} (\ce{H2SO4}) \approx 2$, 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:

$$\ce{C2H5OH + HSO4- &lt;&lt;=&gt; C2H5OH2+ + SO4^2-}\tag{4}$$

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:

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

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

$$\ce{SO4^2-} &gt; \ce{H2O} \approx \ce{C2H5OH} &gt; \ce{HSO4-}\tag{6}$$

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