(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.
H_{2}SO_{4} ----> H^{+} + HSO_{4}^{-}
HSO_{4}^{-} <----> H^{+} + SO_{4}^{-2}
H_{2}SO_{4} | ---> | H^{+} | + | HSO_{4}^{-} | |
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 |
HSO_{4}^{-} | <---> | H^{+} | + | SO_{4}^{-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 [H_{2}SO_{4}] < 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 H_{2}SO_{4}
Ka2 = {[H^{+}][SO_{4}^{-2}]}/[ HSO_{4}^{-}]
1.2 x 10^{-2} = {(0.10 + x)(x)}/(0.10 - x)
0.10x + x^{2} = 1.2 x 10^{-2}(0.10 - x)
x^{2 } + 0.112x - 1.2 x 10^{-3 }= 0
x = (- 0.112 + b^{2} - 4ac)/2a
x = (- 0.112 + 0.0173)/2
x = 0.01 M = [SO_{4}^{-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 SO_{4}^{-2}, and must be added to the 0.10 M H^{+} from the first dissociation of H_{2}SO_{4}.