The ionization constant of HF, HCOOH and HCN at 298K are `6.8 xx 10^(-4), 1.8 xx 10^(-4)` and `4.8 xx 10^(-9)` respectively. Calculate the ionization constants of the corresponding conjugate base.

Explanation

For `F^(Θ), K_(b)=K_(w)//K_(a)=10^(-14)//(6.8xx10^(-4))=1.47xx10^(-11)`
`=1.5xx10^(-11)`.
For `HCOO^(Θ), K_(b)=10^(-14)//(1.8xx10^(-4))=5.6xx10^(-11)`
For `CN^(Θ), K_(b)=10^(-14)//(4.8xx10^(-9))=2.08xx10^(-6)`