An unknown acid that we will call HA ionizes 3.9%. If the concentration of the acid was originally 0.222M...?

An unknown acid that we will call HA ionizes 3.9%. If the concentration of the acid was originally 0.222M, what is the value of Ka?

Answer:
% ionization is [H+]/[HAoriginal], so you can find [H+] from this:

H+ = 0.039*0.222 M = 0.008658 M
This is also equivalent to [A-] at equilibrium.
HA at equilibrium is then [HA] - [H+] = 0.213342 M

Since the dissociation rxn is HA ~> H+ + A-,
The expression for Ka is:

Ka = [H+][A-]/[HA] (all concentration at equilibrium, not the original ones)

Plug in the values and you get Ka = 3.5e-4

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