pH of sulphuric acid calculation

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

If the concentration of the hydrogen ions from the second ionisation only is x, then the concentration of HSO₄^- ions will be 0.1 - x and the total concentration of hydrogen ions from both ionisations will be 0.1 + x. Since it is the total hydrogen ion concentration that appears in the evaluation of K₂, we have:

K₂ =

[H₃O⁺] [SO₄^2-]
[HSO₄^-]

= (0.1 + x) x
(0.1 – x)

= 0.01 mol dm^-3

Thus after a little algebra we obtain

x² + 0.11 x – 0.001 = 0

x =

- 0.11 ± Ö(0.11² + 0.004) =
2

- 0.11 ± 0.1269
2

Since a concentration must be a positive quantity the negative root makes no physical sense; thus

x = 8.45 x 10^-3 mol dm^-3.

The total hydrogen ion concentration is therefore 0.10845 mol dm^-3 which, using a sensible number of significant figures (0.109 mol dm^-3) gives a pH of 0.96.