Given:

pH of Codeine(C₁₈H₂₁NO₃) solution = 9.95

[Cod] = 0.005M

Ionisation of Cod:

Cod + H₂O ↔ CodH⁺ + OH^-1

As we know that pH + pOH = 14

∴ 9.95 + pOH =14

⇒pOH = 14- 9.95

⇒pOH = 4.05

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

∴ 4.05 = -log[OH]

By taking antilog of both the sides, we get

Antilog 4.05 = -[OH^-]

⇒[OH^-] = 8.913 × 10^-5

As we know that, K_b(ionization constant)

⇒K_b=

Hence, for the reaction, Cod + H₂O ↔ CodH⁺ + OH^-1

K_b =

As the reaction is completely ionized,

∴ [CodH⁺] = [OH^-]

Hence, K_b=

As [OH^-] = 8.913 × 10^-5 (calculated)

[Cod] = 0.005M

K_b=

⇒K_b = 1.588 × 10^-6

Thus, the ionization constant of codeine is 1.588 × 10^-6

To calculate Pk_b we apply the formula:

Pk_a = -logK_b

As K_b = 1.74 × 10^-3

∴ pk_b = -log (1.588 × 10^-6)

⇒Pk_b = -log 1.588 – (-6) log 10

⇒Pk_b = 5.80

Thus, the Pk_b of codeine is 5.80