Question

Codeine (C₁₈H₂₁NO₃) is a narcotic pain reliever titrated with HCL. What is the pH of a 0.0500 M codeine hydrochloride solution with a pK_a of 5.80? Note the titrated form of codeine hydrochloride will act as a weak acid due to the formation of soluble ammonium salt.

Answer 1

Since codeine hydrochloride act as acid, in short lets write the symbol as BH+

use:
pKa = -log Ka
5.8 = -log Ka
Ka = 1.585*10^-6

BH+ dissociates as:

BH+          ----->     H+   + B
5*10^-2                 0         0
5*10^-2-x               x         x


Ka = [H+][B]/[BH+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.585*10^-6)*5*10^-2) = 2.815*10^-4

since c is much greater than x, our assumption is correct
so, x = 2.815*10^-4 M



So, [H+] = x = 2.815*10^-4 M


use:
pH = -log [H+]
= -log (2.815*10^-4)
= 3.5505
Answer: 3.55