SELECT * FROM question_mgmt as q WHERE id=3265 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 the ph of 0 005m codeine c18h21no3 solution is 9

Question:

The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

Answer:

c = 0.005

pH = 9.95

pOH = 4.05

pH = -log (4.05)

4.05 =  - log [OH-]

[OH-] = 8.91x 10-5

cα = 8.91x 10-5

α = 8.91x 10-5  /  5 x 10-3  =  1.782  x 10-2

Thus , Kb  =  cα2

= 0.005 x (1.782)2 x 10-4

= 0.005 x 3.1755 x 10-4

= 0.0158 x 10-4

Kb  =  1.58 x 10-6

PKb  =  -logKb 

= -log (1.58 x 10-6)

= 5.80

 


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