The ionization constants of `HF, H COOH` and HCN at 298 K are `6.8x10^(-4), 1.8xx10^(-4)` and `4.7xx10^(-9)` respectively. Calculate the ionization constants of the corresponding conjugate base.

Explanation

(i) Consider the case of HF
`K_(a)=6.8xx10^(-4)`
`K_(b)=(K_(w))/(K_(a)=(1xx10^(-14))/(6.8xx10^(-4))=0.147xx10^(-10)`
(ii) Consider the case of HCOOH
`K_(a)=1.8xx10^(-4)`
`K_(b)=(K_(w))/(K_(a))=(1xx10^(-14))/(1.8xx10^(-4))=0.56xx10^(-10)`
(iii) Consider the case of HCN
`K_(a)=4.8xx10^(-9)`
`K_(a)=(K_(w))//(K_(a))=(1xx10^(-14))/(4.8xx10^(-9))=0.208xx10^(-5)`