Question

The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4 , 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer 1

The relation between ionization constant of an acid and that of its conjugate base is K_a 

x K_b= K_w

K_a x K_b = K_w 

∴ K_b = \(\frac{K_w}{K_a}\)

The conjugate base of HF is F^- , 

∴ K_b(F^-) = \(\frac{K_w}{K_{a(HF)}}\) = \(\frac{1\times10^{-14}}{6.8\times10^{-4}}\) = 1.5 x 10^-11

The conjugate base of HCOOH is HCOO^-.

∴ K_b(HCOO^-) = \(\frac{K_w}{K_{a(HCOOH)}}\) =  \(\frac{1\times10^{-14}}{1.8\times10^{-4}}\)  = 5.6 x 10^-11 

The conjugate base of HCN is CN^-.

∴ K_b(CN-) = \(\frac{K_w}{K_{a(HCN)}}\)  = \(\frac{1\times10^{-14}}{4.8\times10^{-9}}\) = 2.1 x 10^-9