chemistry
posted by ciaociao on .
What is the acid dissociation constant for a 0.010 M solution of the unkown acid HX that has a pH of 2.70 at equilibrium?
I know the answer but not how to get it, please help? Thanks in advance.

pH = 2.70 = log(H^+)
(H^+) = 0.002M
...........HX ==> H^+ + X^
I.......0.010.....0......0
C..........y.....y......y
E........0.010y...y.....y
Ka = (H^+)(X^)/(HX)
y = 0.002 from the pH calculation above. Substitute and solve for Ka. 
Thanks for the help, but I keep getting 4*10^4 and not 5*10^4 which is what I'm told the correct answer is. :S