Calculate the pH of a 3.40×10-4 M solution of codeine if the pKa of the conjugate acid is 8.21.

Question

Codeine is a popular prescription pain killer because it is much less addictive than morphine. Codeine contains a basic nitrogen atom that can be protonated to give the conjugate acid of codeine. Calculate the pH of a 3.40×10^-4 M  solution of codeine if the pK_a of the conjugate acid is 8.21.

Answer 1

Codeine is a base, and can be represented as B. In solution, we have the following equilbrium...

B + H2O <--> BH⁺ + OH^-

From the pKa, we can find the Kb, which we will need since we are dealing with hydrolysis of a base:

pKa + pKb = 14

pKb = 14 - 8.21 = 5.79

Kb = 1x10^-5.79 = 1.62x10^-6

Kb = 1.62x10^-6 = [BH⁺][OH^-] / [B]

1.62x10^-6 = (x)(x) / 3.40x10^-4 - x

x² = 5.51x10^-10 - 1.62x10^-6x

x² + 1.62x10^-6x - 5.51x10^-10 = 0

Use the quadratic formula to solve for x....

x = 2.27x10^-5 M = [OH^-]

pOH = -log [OH^-] = -log 2.27x10^-5 = 4.64

pH = 14 - pOH

pH = 9.356