A0.12 m solution of a monoprotic acid is 2.3% ionized. what is the ka of this acid?

Ionization of an acid of form (HA) can be represented as follows

HA   ↔     H+          +        A-

Ka = 

Now, initial conc. of HA = 0.12 M

Of these, 2.3 % got ionized.

∴ [H+] = [A-] = 0.12 X 0.023 = 0.00276 M

Thus [HA] left after ionization = 0.12 - 0.00276 = 0.11724 M

Now Ka =    = 6.497 x 10^-5

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.

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