The pH of 0.005 M codeine …
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Posted by Pragati Singh 7 years, 10 months ago
- 1 answers
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Gurvinder Kaur 7 years, 10 months ago
From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:
[H+] = antilog (-pH)
= antilog [-9.95]
= 1.12 X 10-10
So, [OH-] = Kw / [H+]
= (1 X 10-14) / (8.91 X 10-5)
= 8.91 X 10-5 M
The concentration of the corresponding codenium ion is also the same as that of hydroxyl ion. Let the codeine ion be denoted by [M+]. Thus
Kb = [M+][OH-] / [MOH]
= (8.91 X 10-5)2 / 0.005
= 1.59 X 10-6
pKb = -log Kb = - log(1.59 X 10-6)
pKb = 5.80
2Thank You