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) + SO42-(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+] [SO42-] [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.

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Dr Rod Beavon   17 Dean's Yard   London  SW1P 3PB

e-mail:  rod.beavon@westminster.org.uk