NOT CORRECT |
(9.a) pH = 7.46
Na_{2}SO_{4} | <----> | 2Na^{+} | + | 2Na^{+} |
0.10 M | 0.10 M |
This salt dissociates completely.
SO_{4}^{-2} | + | H_{2}O | ---> | HSO_{4}^{-} | + | OH^{-} | |
Initial(I) | 0.10 M | 0 M | 1.0 x 10^{-7}M* | ||||
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
x^{2} = 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 [SO_{4}^{-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 = 7.38