A 2.9M methylamine (CH3NH2) solution is 1.3% ionized. What are the pH and Kb values?
Let α be the dissociation of the weak base
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 c = concentration = 2.9 M
Also percentage of ionization = α x 100 = 1.3
α = 1.3x10-2
Kb = cα2 = 2.9 x(1.3x10-2)2
= 4.90x10-4
So the concentration of [OH-] = cα
= 2.9 x1.3x10-2
= 0.0377 M
pOH = - log [OH-]
= - log 0.0377
= 1.42
So pH = 14 - pOH
= 14 - 1.42
= 12.58
A 2.9M methylamine (CH3NH2) solution is 1.3% ionized. What are the pH and Kb values?
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