NOT CORRECT


(9.b) pH = 6.54

This is the value of the pOH NOT the pH

Remember pH + pOH = 14

also....



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 = 6.54 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 = 7.38

Return to Problem Set