SELECT * FROM question_mgmt as q WHERE id=3296 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 a 0 02 m solution of pyridinium hydrochloride has

Question:

A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

Answer:

Given,  pH  =  3.44

We know that

pH  =  -log [H+]

∴ [H+]  =  3.63x10-4

Then Kh  = ( 3.63x10-4 )2 / 0.02         (∵ Concentration is 0.02M)

⇒ Kh  =  6.6 x 10-6

Now , Kh  =  Kw / Ka

⇒  Ka  =  Kw / Kh

= 10-14 / 6.6 x 10-6

= 1.51 x 10-9

 


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