Question

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.

Answer 1

For `F^(Θ), K_(b)=K_(w)//K_(a)=10^(-14)//(6.8xx10^(-4))=1.47xx10^(-11)`
`=1.5xx10^(-11)`.
For `HCOO^(Θ), K_(b)=10^(-14)//(1.8xx10^(-4))=5.6xx10^(-11)`
For `CN^(Θ), K_(b)=10^(-14)//(4.8xx10^(-9))=2.08xx10^(-6)`