# The pH of 0.005M

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, Kb(ionization constant)

Kb= Hence, for the reaction, Cod + H2O CodH+ + OH-1

Kb = As the reaction is completely ionized,

[CodH+] = [OH-]

Hence, Kb= As [OH-] = 8.913 × 10-5 (calculated)

[Cod] = 0.005M

Kb= Kb = 1.588 × 10-6

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

To calculate Pkb we apply the formula:

Pka = -logKb

As Kb = 1.74 × 10-3

pkb = -log (1.588 × 10-6)

Pkb = -log 1.588 – (-6) log 10

Pkb = 5.80

Thus, the Pkb of codeine is 5.80

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