Given:

[C₆H₅NH₂] = 0.001M

Ionization constant of aniline (K_a) = 4.27 × 10^-10

Ionisation of aniline:

C₆H₅NH₂ + H₂O ↔ C₆H₅NH₂ + OH^-

K_a =

At equilibrium, []

∴ K_a =

As K_a = 4.27 × 10^-10 (given)

[C₆H₅NH₂] = 0.001M (given)

∴ 4.27 × 10^-10 = .

⇒[OH] = √(4.27 ×0^-10) (0.001)

⇒[OH] = 6.534 × 10^-7

As we know that, pOH = -log[OH^-]

∴ pOH = -log[OH]

⇒pOH = -log (6.534 × 10^-7)

⇒pOH = 6.18

As we know that pH + pOH = 14

∴ pH + 6.18 =14

⇒pH = 14- 6.18

⇒pH = 7.82

By applying the formula

K_b =

⇒K_b =

⇒K_a=

We can neglect C(1-α) as its value is very small

∴ K_b = Cα²

As C = 0.001 M (Given)

K_b = 4.27 × 10^-10 (given)

Hence,

⇒4.27 × 10^-10= 0.001× α²

⇒α = 6.53 × 10^-4

Thus, the degree of ionization aniline is 6.53 × 10^-4

As we know that pk_a + pK_b = 14 (for a conjugate acid and base)

∴ Pk_a = -logK_b

As K_b = 4.27 × 10^-10 (given)

∴ pk_a = -log ( 4.27 × 10^-10 )

⇒Pk_a= -log 4.27 – (-10) log 10

⇒Pk_a = -0.62 + 10

⇒Pk_a = 9.37

pk_a + pK_b = 14

⇒Pk_a = 14 – 9.37

⇒Pk_a = 4.63

As we know that, pk_a = -logK_a

-log K_a = 4.63

By taking antilog of both the sides, we get

K_a = antilog 4.63

⇒K_a = 2.4 × 10^-5

Thus, the ionization constant of the conjugate acid of aniline is 2.4 × 10^-5