SELECT * FROM question_mgmt as q WHERE id=3266 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) CBSE Free NCERT Solution of 11th chemistry Equilibrium what is the ph of 0 001 m aniline solution the io

Question:

What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer:

Kb  =   4.27 x 10-10

c  =  0.001M

pH  = ?

α  =  ?

Kb  =  cα2

4.27 x 10-10  = 0.001 x α2

4270 x 10-10 = α2

α = 65.34 x 10-4

Then (anion) = cα  = 0.001 x 65.34 x 10-4

= 0.65 x 10-5

pOH = -log ( 0.65 x 10-5)

= 6.187

pH  =  7.813

Now

Ka x Kb = Kw

 ∴ 4.27 x 10-10 x Ka = Kw

Ka  =  10-14 / 4.27 x 10-10

= 2.34 x 10-5

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10-5. 


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