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?

Question

I know the answer but not how to get it, please help? Thanks in advance.

Answer 1

pH = 2.70 = -log(H^+)
(H^+) = 0.002M

...........HX ==> H^+ + X^-
I.......0.010.....0......0
C..........-y.....y......y
E........0.010-y...y.....y

Ka = (H^+)(X^-)/(HX)
y = 0.002 from the pH calculation above. Substitute and solve for Ka.

Answer 2

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