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Question

The ionization constant of HF, HCOOH and HCN at 298K are 6.8×104,1.8×104 and 4.8×109 respectively. Calculate the ionization constants of the corresponding conjugate base. 


Solution

It is known that,

Kb=KwKa
Given,
Ka of HF= 6.8×104
Hence, Kb of its conjugate base F
=KwKa=10146.8×104=1.5×1011
Guven,
Ka of HCOOH=1.8×104
Hence, Kb of its conjugate base HCOO
=KwKa=10141.8×104=5.6×1011
Guven,
Ka of HCN=4.8×109
Hence, Kb of its conjugate base CN
=KwKa=10144.8×109=2.08×106


 

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