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 constants of the corresponding conjugate base.

The ionization constant of HF, HCOOH and HCN at 298 K are 6.8xx10^(-4)

1.	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 constants of the corresponding conjugate base.
Answer» SOLUTION :`F^-` is a conjugate base of HF, its ionization constant `K_b` is, `K_b=K_w/K_a=(1.0xx10^(-14))/(6.8xx10^(-4))= 1.47xx10^(-11)` `HCOO^-` is a conjugate base of HCOOH , its ionization constant `K_b`, `K_b=K_w/K_a=(1.0xx10^(-14))/(1.8xx10^(-4))=5.556xx10^(-11)` `CN^-` is a conjugate base of HCN, its ionization constant `K_b` is , `K_b=K_w/K_a=(1.0xx10^(-14))/(4.8xx10^(-9))= 2.08xx10^(-6)`