Question

A 0.040M solution of a monoprotic acid is 14% ionized. Calculate the Ka for the weak acid.

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so:
HX <---> H+ + X-

Ka= [H+][X-]/ [HX]

Since its a monoprotic acis i know the concentration of H and X will be equal

And the HX will be 86% (100%-14%)

BUt i m not sure how to solve for it

Answer 1

So (H^+) = 0.04 x 0.14 = ??
(X^-) is the same.
(HX) = either,
0.04-(0.04*0.14) OR
0.04*0.86 (HX) will be the same no matter which way you go.