A 1.0 x 10-2 M solution of a weak acid is found to be 3.7% dissociated. Calculate Ka of this acid.

Question

I found the amount dissociated by (amount dissociated/initial conc)*100=3.7
Where do I go from here?

Answer 1

I think you need to back up before going forward.
(HA) = 0.01
(H^+) = (A^-) = 0.01 x 0.037 = 3.7E-4

............HA ==> H^+ + A^-
I.........0.01.....0.....0
C........-3.7E-4..3.7E-4..3.7E-4
E.......you finish

Substitute the E line into the Ka expression and solve for Ka. Your answer should be approx 1E-5 if I plugged in the right numbers.