(9.d) pH = 7.39
| Na2SO4 | <----> | 2Na+ | + | 2Na+ |
| 0.10 M | 0.10 M |
This salt dissociates completely.
| SO4-2 | + | H2O | ---> | HSO4- | + | OH- | |
| Initial(I) | 0.10 M | 0 M | 1.0 x 10-7M* | ||||
| Change(C) | -x | + x | +x | ||||
| Equilibrium(E) | (0.10-x) | x | 1.0 x 10-7 + x |
* Kw = (Ka)(Kb)
1 x 10-14 = (1.2 x 10-2) Kb
Kb = 8.3 x 10-13 and since Kb is not 100 times larger than Kw the [OH- ] in pure water must be considered and ...
8.3 x 10-13 = {(x)(x)}/0.10
x2 = 8.3 x 10-14
x = 2.9 x 10-7 (this is not a valid approximation since x should be at least 10 times larger than 1. 0 x 10-7 and pOH = 6.5 and pH = 7.46 are not correct).
So ...
8.3 x 10-13 = {x (1.0 x 10-7) + x}/(0.10 - x)**
** Since [SO4-2 ] > 100 Ka and obeys the 5 percent rule, then (0.10 - x) is ~ 0.10.
Therefore....
8.3 x 10-13 = {x (1.0 x 10-7) + x}/0.10
Use the quadratic equation
a = 1
b = 1.0 x 10-7
c = 8.3 x 10-14
= {1.0 x 10-7 + 5.85 x 10-7}/2
= 2.424 x 10-7
pOH = 6.62 and pH = 7.39