Gchem question: help please

[vindicate]

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Calculate the pH of a 2.0M C2H5NH2 solution at 25 C. Kb is 6x10^-4

I can get the answer if I solve like below:
6x10^-4 = [x][x]/2.0
x = 3.5x10^-2
-log(3.5x10^-2) = pOH
= 2-.8+.3 = 1.5
pH = 12.5

I can't get it if I solve like:
-log(6x10^-4) = pKb = 4-.47-.3 = 3.2
pKa+pKb=14
pH = 1/2(pKa)+1/2log(HA)
= 1/2(11.8)+1/2log(2.0M)
= 5.9+.15


Can anyone tell me what I'm doing wrong in the latter method that's not giving me the right answer?

 

[CANgnome]

isn't HH:

pH = log pKa + log [A-]/[HA]?

or pOH = log pKb + log [HA+]/[A]?


you aren't even given the concentration of the conjugate acid, why are you plugging it into the equation???