NOT CORRECT |
(7.a) 0.010 M
This is the [H+] is due to the dissociation of the second H+ alone.
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
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= 0.01 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.