Question

A 0.12 m solution of a monoprotic acid is 2.3% ionized. what is the ka of this acid?

Answer 1

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

HA ↔ H+ + A-

Ka =
`([H+][A-])/([HA])`

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 =
`((0.00276 )^2)/(0.11724)` = 6.497 x 10^-5