Question 43

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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