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.

Asked by Pragya Singh | 1 year ago |  85

##### Solution :-

For F$$K_{b} = \dfrac{K_{w}}{K_{a}}=\dfrac{10^{-14}}{(6.8\times 10^{-4})}$$

$$=1.47\times 10^{-11}$$

For HCOO,$$K_{b}=\dfrac{10^{-14}}{(1.8\times 10^{-4})}$$

$$=5.6\times 10^{-11}$$

For CN = $$K_{b}=\dfrac{10^{-14}}{(4.8\times 10^{-9})}$$

$$=2.08\times 10^{-6}$$

Answered by Abhisek | 1 year ago

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