[specialk08]

NaF -> Na+ + F-

F- + H2O    HF + OH-
.10 - x                 +x    +x

So to determine Kb, I have the value of Kw and Ka (Ka for Fluoride = 6.0 x 10^-4)

Kb = Kw/Ka
Kb= 1.10x10^-14 / 6x10^-4
Kb = 6.6x10^-18

Kb = x^2
       0.10 - x
Approximation:
Kb = x^2
       .10

6.6x10^-18 = x^2
                    0.10
So cross multiply:
6.6x10^-19 = x^2
Take sq rt of both sides
x = 8.12x10^-10

[OH-] = 8.12x10^-10 and using that pOH = -log[OH]:

pOH = -log[8.12x10^-10]
pOH = -(-9.09)
pOH = 9.09

pOH + pH = 14
9.09 + pH = 14
pH = 14 -9.09
pH = 4.91

Wow! this one is absolutely right.. Thanks

The math in that is wrong...I suggest double checking it :-). Check out the Kb value and recalculate it.