A 0.010 M solution of a weak monoprotic acid is 3.0% dissociated. What is the equilibrium constant, Ka, for this acid?

Answer 1

HA = H^+ + A^-

Ka = (H^+)(A^-)/(HA)

initial:
HA = 0.01
H^+ = 0
A^- = 0

change:
H^+ = x
A^- = x
HA = 0.01 - x

equilibrium:
x = 0.01 x 0.03 = ??
Substitute and solve for Ka.

Answer 2

3.0 * 10^-5