Question

The hydronium ion concentration of an aqueous solution of 0.538 M methylamine (a weak base with the formula CH₃NH₂) is

[H₃O⁺] = ________ M

Answer 1

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 Kb = 4.4x10^-4
c = concentration = 0.538 M
Plug the values we get α = 0.0286
So the concentration of [OH-] = cα
= 0.0154 M
pOH = - log [OH-]
= 1.81

pH=14-pOH=12.19

-log[H3O+]=12.19

[H3O+]=10^-12.19=6.46×10^-13 M