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

Explanation

We know, `K_(a)xxK_(b)=10^(-14)`
For the conjugate acid-base pair, HF and `F^(-)`
`K_(b)(F^(-))=(10^(-14))/(K_(a)(HF))=(10^(-14))/(6.8xx10^(-4))=1.471xx10^(-11)`
For the conjugate acid-base pair, HCOOH and `HCOO^(-)`
`K_(b)(HCOO^(-))=(10^(-14))/(K_(a)(HCOOH))=(10^(-14))/(1.8xx10^(-4))=5.55xx10^(-11)`
For the conjugate acid-base pair, HCN and `CN^(-)`
`K_(b)(CN^(-))=(10^(-14))/(K_(a)(HCN))=(10^(-14))/(4.8xx10^(-9))=2.08xx10^(-6)`.