Given:

The ionization constant of HF = 6.8 × 10^–4

The ionization constant of HCOOH = 1.8 × 10^-4

The ionization constant of HCN = 4.8 × 10^-9

⇒Conjugate base of HF = F^-

⇒Conjugate base of HCOOH = HCOO^-

⇒Conjugate base of HCN = CN^-

To find out the ionization constants of the corresponding conjugate base, we apply the formula:

K_w = K_a × K_b

Where K_w is the ionic productof water.

K_a is the ionization constant of acid

K_b is the ionization constant of base

To calculate the ionization constant of conjugate base of HF, i.e., F^-,

Using the formula, we write

K_b =

As K_a of HF = 6.8 × 10^–4 (given)

K_w = 10^-14 (same value for every acid or base)

⇒K_b =

⇒K_b = 1.5 × 10^-11

To calculate the ionization constant of conjugate base of HCOOH, i.e., HCOO^-,

Using the formula, we write

K_b =

As K_a of HCOOH = 1.8 × 10^–4 (given)

K_w = 10^-14

⇒K_b =

⇒K_b = 5.6 × 10^-11

To calculate the ionization constant of conjugate base of HCN, i.e., CN^-,

Using the formula, we write

K_b =

As K_a of HCN = 4.8 × 10^–9 (given)

K_w = 10^-14

⇒K_b =

⇒K_b = 2.08 × 10^-6