Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:
C5H5N+H2O⇌C5H5NH++OH−
The pKb of pyridine is 8.75. What is the pH of a 0.245 M solution of pyridine?
Express the pH numerically to two decimal places.
use:
pKb = -log Kb
8.75= -log Kb
Kb = 1.778*10^-9
C5H5N dissociates as:
C5H5N +H2O -----> C5H5NH+ + OH-
0.245 0 0
0.245-x x x
Kb = [C5H5NH+][OH-]/[C5H5N]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.778*10^-9)*0.245) = 2.087*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.087*10^-5 M
So, [OH-] = x = 2.087*10^-5 M
use:
pOH = -log [OH-]
= -log (2.087*10^-5)
= 4.6804
use:
PH = 14 - pOH
= 14 - 4.6804
= 9.3196
Answer: 9.32
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