This seemingly simple question is one that has been incorrectly treated in books and even in examination questions. The simple view is that because sulphuric acid is dibasic the concentration of hydrogen ions in the solution must be 0.2 mol dm^-3, which gives a pH of 0.7. This is wrong.

The incompleteness of the second ionisation of sulphuric acid is crucial; it is heavily suppressed by the first ionisation, which is complete. This is a simple application of le Chatelier's Principle - at least it is when you come to think of it. The result is that the pH of the 0.1 mol dm^-3 acid is about 0.98. If sulphuric acid were monobasic the expected pH would be 1; the second ionisation makes very little difference.

Sulphuric acid behaves essentially as a monobasic acid as far as the pH of the solution is concerned.

Consider a 0.1 mol dm^-3 solution of sulphuric acid. The equilibria in this solution are:

H₂SO₄(aq) + H₂O(l) à H₃O⁺(aq) + HSO₄^- (aq)       K₁ = very large

H₂O(l) + HSO₄^- (aq)  Û   H₃O⁺(aq) + SO₄^2-(aq)      K₂  = 0.01 mol dm^-3

The first ionisation is complete so the concentration of hydrogen ions from this is 0.1 mol dm^-3.

A 0.1 mol dm^-3. solution of sodium hydrogen sulphate NaHSO₄ which contains only the second ionisation shown above has a pH of 1.57. This corresponds to a hydrogen ion concentration of 0.027 mol dm^-3. Thus it seems that 0.1 mol dm^-3 sulphuric acid should have a hydrogen ion concentration of (0.1 + 0.027) = 0.127 mol dm^-3, this being the sum of each concentration taken separately.

In fact the hydrogen ion concentration is only 0.105 mol dm^-3 (pH = 0.98), showing the effect of the first ionisation in suppressing the second.

This argument is taken from the measured pH of the various solutions. A similar result is obtained if you calculate the hydrogen ion concentrations from the values of K_a from the two equilibria. Try it; if you get stuck my answer is here.