LIVE Course for free

Rated by 1 million+ students
Get app now
+1 vote
in Chemistry by (29.8k points)

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. 

1 Answer

+1 vote
by (128k points)
selected by
Best answer

It is known that, 


Ka of HF = 6.8 × 10–4 

Hence, Kb of its conjugate base F– 


Ka of HCOOH = 1.8 × 10–4 

Hence, Kb of its conjugate base HCOO– 


Ka of HCN = 4.8 × 10–9 

Hence, Kb of its conjugate base CN–

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.