What is the dissociation reaction for this solution?
Consider a 0.35 M pyridine solution, C5H5N(aq), that has a pH of 9.38.
1)
C5H5N (aq) + H2O (l)
<-> C5H5NH+(aq)
+ OH-(aq)
2)
use:
pH = -log [H+]
9.38 = -log [H+]
[H+] = 4.169*10^-10 M
use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at
25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(4.169*10^-10)
[OH-] = 2.399*10^-5 M
C5H5N dissociates as:
C5H5N +H2O
-----> C5H5NH+ +
OH-
0.35
0 0
0.35-x
x x
Kb = [C5H5NH+][OH-]/[C5H5N]
Kb = x*x/(c-x)
Kb = 2.399*10^-5*2.399*10^-5/(0.35-2.399*10^-5)
Kb = 1.644*10^-9
Answer: 1.6*10^-9
What is the dissociation reaction for this solution? Consider a 0.35 M pyridine solution, C5H5N(aq), that...
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