1.

The ionisation constant of HF, HCOOH ,HCN at 298 K are 6.8 times 10^-4.1.8 times 10^-4 and 4.8 times 10^-9. Respectively. Calculate the ionisation constant of the corresponding conjugate base.

Answer»

Solution :(i) HF, CONJUGATE base F
`K_b=K_w//K_a=(1 times 10^-4)/(6.8 times 10^-4)=1.47 times 10^-11= 1.5times 10^-11`
(ii) for `HCOO^-`
`K_b=(1 times 10^-14)/(1.8 times 10^-4)=5.6 times 10^-11`
(iii) for `CN^-`
`K_b=(1 times 10^-14)/(4.8 times 10^-4)=2.8 times 10^-6`


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