SELECT * FROM question_mgmt as q WHERE id=3077 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 ionization constant of hf hcooh and hcn at 29

Question:

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4,  1.8 × 10–4  and  4.8 × 10–9 respectively.

Calculate the ionization constants of the corresponding conjugate base.

Answer:

It is known that,

Kb  =  Kw / Ka

Given

Ka of HF = 6.8 × 10-4

Hence, Kb of its conjugate base F-  =  Kw / Ka

= 10-14 / 6.8 × 10-4

= 1.5 x 10-11

Given, Ka of HCOOH = 1.8 × 10-4

Hence, Kb of its conjugate base HCOO-  = Kw / Ka

= 10-14 / 1.8 × 10-4

= 5.6x10-11

Given, Ka of HCN = 4.8 × 10-9

Hence, Kb of its conjugate base CN- = Kw / Ka

= 10-14 / 4.8 × 10-9

= 2.8 x 10-6


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