Question

Ethanolamine,HOC2H4NH2 , is a viscous liquid with an ammonia-like odor; it is used to remove hydrogen sulfide from natural gas. A 0.60 M aqueous solution of ethanolamine has a pH of 11.60. What is Kb for ethanolamine? Calculate the pH of a 0.41 M aqueous solution of ethanolamine

Answer 1

a)

pH = 11.60

pOH = 14 - pH

= 14 -11.60

= 2.4

USE:

pOH = -log [OH-]

2.4 =- log [OH-]

[OH-] = 3.98*10^-3 M

for simplicity lets write ethanolamine as ANH2

ANH2 + H2O   ——> ANH3+       +    OH-

0.60 0 0 (initial)

0.60-x                   x               x       (at equilibrium)

[OH-] = x = 3.98*10^-3 M

Kb = [ANH3+][OH-]/[ANH2]

Kb = x*x/(0.60-x)

= (3.98*10^-3 * 3.98*10^-3 ) / (0.60 - 3.98*10^-3)

= 2.66*10^-5

b)

Kb = x*x/(0.41-x)

since Kb is small, x will be small and can be ignored as compared to 0.41

2.66*10^-5 = x^2/0.41

x = 3.3*10^-3 M

so,

[OH-] = 3.3*10^-3 M

pOH = -log [OH-]

= -log(3.3*10^-3 )

= 2.48

pH = 14 - pOH

= 14 - 2.48

=11.52

Answer: 11.52