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 1

It is known that, 

Given, 

Ka of HF = 6.8 × 10–4 

Hence, Kb of its conjugate base F– 

Given, 

Ka of HCOOH = 1.8 × 10–4 

Hence, Kb of its conjugate base HCOO– 

Given, 

Ka of HCN = 4.8 × 10–9 

Hence, Kb of its conjugate base CN–