Question

Question 7.43 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.

Class XI Equilibrium Page 228

Answer 1

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Answer 2

we know, if Ka is the Ionisation constant of weak acid ( HA) and Kb is the Ionisation constant of it's conjugate base (A-)
then, Ka.Kb = Kw
given, Ka of HF is 6.8 × 10⁻⁴
then,Kb = Kw/Ka
= 10⁻¹⁴ /6.8 × 10⁻⁴
= 1.47 × 10⁻¹¹

Ka of HCOOH is 1.8 × 10⁻⁴
then, Kb = Kw/Ka
= 10⁻¹⁴ /1.8 × 10⁻⁴
= 5.56 × 10⁻¹¹

Ka of HCN is 4.8 × 10⁻⁹
then, Kb = Kw/Ka
= 10⁻¹⁴/4.8 × 10⁻⁹
= 2.08 × 10⁻⁶