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

(i)`K_a(HF)=6.8xx10^(-4)`
`K_b(F^-)=K_w/K_a =(1xx10^(-14))/(6.8xx10^(-4))=1.47xx10^(-11)`
(ii)`K_a (HCOOH)=1.8xx10^(-4)`
`K_b(HCOO^-)=(1xx10^(-14))/(1.8xx10^(-4))=5.6xx10^(-11)`
(iii)`K_a (HCN)=4.8xx10^(-9)`
`K_b(CN^-) =(1xx10^(-14))/(4.8xx10^(-9))=2.08xx10^(-6)`