The Total solution for NCERT class 6-12
The ionization constant of HF, HCOOH and HCNat 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.
Itis known that,
Given,
Ka of HF = 6.8 × 10–4
Hence, Kb of its conjugatebase F–
Ka of HCOOH = 1.8 × 10–4
Hence, Kb of its conjugatebase HCOO–
Ka of HCN = 4.8 × 10–9
Hence, Kb of its conjugatebase CN–