Question

#2 Ethylamine (CH₃CH₂NH₂) is a base compound.

a) Write base dissociation reaction ethylamine.

b) If K_b of ethylamine is 7.41x10^-10 , what is the pH of 0.15 M ethylamine solution?

Answer 1

a)

CH3CH2NH2 +H2O -----> CH3CH2NH3+ + OH-

b)

Lets write the dissociation equation of CH3CH2NH2

CH3CH2NH2 +H2O -----> CH3CH2NH3+ + OH-

0.15 0 0

0.15-x x x

Kb = [CH3CH2NH3+][OH-]/[CH3CH2NH2]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((7.41*10^-10)*0.15) = 1.054*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.054*10^-5 M

So, [OH-] = x = 1.054*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.054*10^-5)

= 4.98

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.98

= 9.02

Answer: 9.02