Sulfuric acid is a diprotic acid, with the special characteristic of presenting a first complete dissociation. The pH of a sulfuric acid solution is given, approximately, by the protons generated in the first dissociation since the second presents little extension.

Let's see the ionization reactions of sulfuric acid:

$H_2SO_4 +H_2O\rightarrow HSO_4^- +H_3O^+$

$HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+$ $k_a=1.1x10^{-2}$

As can be seen, the first reaction is completely shifted to the right and its equilibrium constant is not shown since it is very large. On the contrary, the second reaction does not totally displace and we must evaluate its equilibrium as a function of the constant, as well as the protons generated in the first reaction.

Let us now calculate the pH of a 0.5 M solution of sulfuric acid.

$H_2SO_4 +H_2O\rightarrow HSO_4^- +H_3O^+$

--- --- 0.5 0.5

$HSO_4^-+H_2O\rightleftharpoons SO_4^{2-}+H_3O^+$

0.5 - x x 0.5 + x

We evaluate the constant of this second equilibrium:

\begin{equation}K_{a}=\frac{[SO_4^{2-}][H_3O^+]}{[HSO_4^-]}=\frac{x(0.5+x)}{0.5-x} =1.1x10^{-2}\end{equation}

Since x<<0.5, we can neglect it in the sums, we are left with:

\begin{equation}1.1x10^{-2}=\frac{0.5x}{0.5}\end{equation}

Equation from which it is easily obtained, $x=0.011\;M$

Therefore, the total proton concentration in the medium is: $[H_3O^+]=0.5+0.011=0.511\;M$