SELECT * FROM question_mgmt as q WHERE id=3293 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

The ionization constant of propanoic acid is 1.32 x 10-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answer:

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA     +    H2O      ↔      H3O+     +  A-

(0.05-0.0α) ≈ 0.05       0.05α     0.05α

Ka  =  [H3O+]  [A-]  / [HA]

=  (0.05α)(0.05α)  /  0.05

= 0.05 α2

Then, [H3O+]  = 0.05α = 0.05 x 1.63 x 10-2  =  Kb . 15 x 10-4 M

∴ pH  =  3.09

In the presence of 0.1M of HCl, let α' be the degree of ionization.

Then, [H3O+]  = 0.01

[A-]  =  0.05α'

[HA]  =  0.05

Ka  =  0.01 x 0.05α'  / 0.05

⇒ 1.32 x 10-5 =  0.1 α'

α' =  1.32 x 10-3


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