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  • 14-05-2018
  • Chemistry
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A 0.12 m solution of a monoprotic acid is 2.3% ionized. what is the ka of this acid?

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PBCHEM
PBCHEM PBCHEM
  • 23-05-2018
Ionization of an acid of form (HA) can be represented as follows

HA   ↔     H+          +        A-

Ka = [tex] \frac{[H+][A-]}{[HA]} [/tex]

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 =  [tex] \frac{(0.00276 )^2}{0.11724} [/tex]  = 6.497 x 10^-5
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