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.
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....
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