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.
It is known that,
Ka of HF=6.8×10-4
Ha of HCOOH=1.8×10-4Hence, Kb of its conjugate base HCOO-=KwKa=10-141.8×10-4=5.6×10-11
Ka of HCN=4.8×10-9Hence, Kb of its conjugate base CN-=KwKa=10-144.8×10-9=2.08×10-6
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