In: Chemistry
what is the pH of 0.124 M pyridine (C5H5N)?
Let α be the dissociation of the weak base,pyridine
BOH
<---> B + + OH-
initial conc. c 0 0
change -cα +cα +cα
Equb. conc. c(1-α) cα cα
Dissociation constant, Kb = (cα x cα) / ( c(1-α)
= c α2 / (1-α)
In the case of weak bases α is very small so 1-α is taken as 1
So Kb = cα2
==> α = √ ( Kb / c )
Given Kb = 1.7x10-9
c = concentration = 0.124 M
Plug the values we get α = 1.17x10-4
So the concentration of [OH-] = cα
= 0.124 x1.17x10-4
= 1.45 x 10-5 M
pOH = - log [OH-]
= - log (1.45 x 10-5)
= 4.84
So pH = 14 - pOH
= 14 - 4.84
= 9.16