(7.b) 0.11 M
The [H+] is due primarily to the dissociation of the first H+. There is a two step dissociation. The first has a very large Ka (a strong acid) the second has a smaller Ka and must be found in tables. This turns out to be a "common ion" problem.
H2SO4 ----> H+ + HSO4-
HSO4- <----> H+ + SO4-2
| H2SO4 | ---> | H+ | + | HSO4- | |
| Initial(I) | 0.10 M | 0 M | 0 M | ||
| Change(C) | -0.10 M | +0.10M | +0.10 M |
| Equilibrium(E) | 0 M | 0.10M | 0.10 M |
| HSO4- | <---> | H+ | + | SO4-2 | |
| Initial(I) | 0.10 M | 0.10 M** | 0 M | ||
| Change(C) | -x | +x | +x |
| Equilibrium (E) | (0.10-x)* | (0.10+x)* | x |
*Hint : Since [H2SO4] < 100 Ka2 then x is probably not much smaller than 0.10 so (0.10 - x) and (0.10 + x) must be used in the calculation rather than assuming that (0.10 - x) ~ 0.10.
** from the first dissociation of H2SO4
Ka2 = {[H+][SO4-2]}/[ HSO4-]
1.2 x 10-2 = {(0.10 + x)(x)}/(0.10 - x)
0.10x + x2 = 1.2 x 10-2(0.10 - x)
x2 + 0.112x - 1.2 x 10-3 = 0
x = (- 0.112 + b2 - 4ac)/2a
x = (- 0.112 + 0.0173)/2
x = 0.01 M = [SO4-2]
OR.........
x = (0.112 + 0.132)/2
x = - 0.112(not valid)
This is the contribution of H+ from the 2nd dissociation which forms SO4-2, and must be added to the 0.10 M H+ from the first dissociation of H2SO4.