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 constants of the corresponding conjugate base.

Explanation

`F^-` is a conjugate base of HF, its ionization constant `K_b` is,
`K_b=K_w/K_a=(1.0xx10^(-14))/(6.8xx10^(-4))= 1.47xx10^(-11)`
`HCOO^-` is a conjugate base of HCOOH , its ionization constant `K_b`,
`K_b=K_w/K_a=(1.0xx10^(-14))/(1.8xx10^(-4))=5.556xx10^(-11)`
`CN^-` is a conjugate base of HCN, its ionization constant `K_b` is ,
`K_b=K_w/K_a=(1.0xx10^(-14))/(4.8xx10^(-9))= 2.08xx10^(-6)`