Calculate the pH of a 2.25×10-4 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 2.25×10^-4 M  solution of codeine if the pK_a of the conjugate acid is 8.21.

Answer 1

Codeine acts as a base, so we can let codeine be represented as B, and then write the hydrolysis reaction as follows:

B + H2O <==> BH+ + OH-

Next, we can write the Kb expression:

Kb = [BH+][OH-] / [B]

Since we know the pKa of the conjugate acid, we can find the Kb for the base (codeine):

pKa = 8.21

Ka = 1x10^-8.21 = 6.17x10^-9

KaKb = Kw

Kb = Kw/Ka = 1x10^-14/6.17x10^-9 = 1.62x10^-6

Substituting values into the Kb expression, we have...

Kb = [BH+][OH-] / [B]

1.62x10^-6 = (x)(x) / 2.25x10^-4 - x (b/c Ka is small, we can ignore x in denominator)

1.62x10^-6 = (x)(x) / 2.25x10^-4

x² = 3.65x10^-10

x = 1.91x10^-5 M = [OH-]

pOH = -log 1.91x10^-5 = 4.72

pH = 14 - pOH = 14 - 4.72

pH = 9.28

NOTE: If you are required to find a more accurate answer, then go back and do not ignore x in the denominator. Use the quadratic equation to solve for [OH-] and then find pH from there.