This is what I have done so far but Im not sure if Im on the right track
The molarity of the H2SO4 initially is 0.21g/98mols
= 0.00214 and that in 0.25 L = 0.00285714285 = 0.0086 mols/L.
The first ionization will result in H^+ of 0.0086 mols/L.
Consider a 0.1 mol dm-3 solution of sulphuric acid. The equilibria in this solution are:
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4- (aq) K1 = very large
H2O(l) + HSO4- (aq) H3O+(aq) + SO42-(aq) K2 = 0.01 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
Consider a 0.1 mol dm-3 solution of sulphuric acid. The equilibria in this solution are:
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4- (aq) K1 = very large
H2O(aq) + HSO4- (aq) 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-] = (0.1 + x) x = 0.01 mol dm-3
[HSO4-] (0.1 – x)
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 = (0.109 mol dm-3)
PH of 0.96
