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