Question

If 99.0 of the weak acid stays un-ionized in 1.0 M aqueous solution, what is the
Ka

Answer 1

99% remains as HX, so 1% goes to H+ and X-.

1.0 M * 0.01 = 0.01 M. H+ and X-

1.0 -0.01 = 0.99 M remains as HX.

$HX <---> H+ + X-$

$Ka = [H+][X-] / [HX]$ = (0.01)^2 / 0.99 = 1.01* 10