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

We know that :`K_(a) xx K_(b) =K_(w).`
`"For HCOOH :" " " K_(a) = 6.8 xx 10^(-4) , K_(b) = (K_(w))/(K_(a)) =(10^(-14))/(6.8 xx 10^(-4)) = 1.5 xx 10^(-11)`
`"For HCOOH: "" "K_(a) =1.8 xx10^(-4) , K_(b) =(K_(w))/(K_(a)) =(10^(-14))/(1.8 xx10^(-4)) =5.6xx10^(-11)`
`" For HCN : "" "K_(a)= 4.8 xx 10^(-9) , K_(b) =(K_(w))/(K_(a)) =(10^(-14))/(4.8xx10^(-4))= 2.08 xx 10^(-6).`