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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 43
Total: 73
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Solution:  
Kb =
Kw
Ka
For F− , Kb =
10−14
6.8×10−4
 = 1.47 × 10−11
For HCOO− , Kb =
10−14
1.8×10−4
 = 5.6 × 10−11
For CN− , Kb =
10−14
4.8×10−9
 = 2.08 × 10−6
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