The ionization constant of `HF,HCOOH` and `HCN` at `298 K` are `6.8xx10^(-4), 1.8xx10^(-4)` and `4.8xx10^(-9)` respectively. Calculate the ionization constant of the corresponding conjugate base.

Explanation

For conjugate acid-base pair:
`K_(a)xxK_(b)=K_(w)`
`K_(b)` of `F^(-)= (10^(-14))/(6.8xx10^(-4))=1.47xx10^(-11)`
`K_(b) of HCOO^(-)= (10^(-14))/(1.8xx10^(-4))=5.6xx10^(-11)`
`K_(b)of CN^(-)= (10^(-14))/(4.8xx10^(-8))=2.08xx10^(-7)`