The ionization constant of HF, HCOOH and | Class 11 Chemistry Chapter Equilibrium, Equilibrium NCERT Solutions

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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 43: The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4,  1.8 × 1....