Question

a. Calculate the pH of a 0.538 M aqueous solution of ethylamine (C₂H₅NH₂, K_b = 4.3×10^-4)

b. Calculate the pH of a 0.0473 M aqueous solution of piperidine (C₅H₁₁N, K_b = 1.3×10^-3).

Answer 1

a)

C2H5NH2 dissociates as:

C2H5NH2 +H2O -----> C2H5NH3+ + OH-

0.538 0 0

0.538-x x x

Kb = [C2H5NH3+][OH-]/[C2H5NH2]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((4.3*10^-4)*0.538) = 1.521*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

4.3*10^-4 = x^2/(0.538-x)

2.313*10^-4 - 4.3*10^-4 *x = x^2

x^2 + 4.3*10^-4 *x-2.313*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 4.3*10^-4

c = -2.313*10^-4

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 9.255*10^-4

roots are :

x = 1.5*10^-2 and x = -1.543*10^-2

since x can't be negative, the possible value of x is

x = 1.5*10^-2

So, [OH-] = x = 1.5*10^-2 M

use:

pOH = -log [OH-]

= -log (1.5*10^-2)

= 1.824

use:

PH = 14 - pOH

= 14 - 1.824

= 12.176

Answer: 12.18

b)

C5H11N dissociates as:

C5H11N +H2O -----> C5H11NH+ + OH-

4.73*10^-2 0 0

4.73*10^-2-x x x

Kb = [C5H11NH+][OH-]/[C5H11N]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.3*10^-3)*4.73*10^-2) = 7.842*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

1.3*10^-3 = x^2/(4.73*10^-2-x)

6.149*10^-5 - 1.3*10^-3 *x = x^2

x^2 + 1.3*10^-3 *x-6.149*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.3*10^-3

c = -6.149*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 2.476*10^-4

roots are :

x = 7.218*10^-3 and x = -8.518*10^-3

since x can't be negative, the possible value of x is

x = 7.218*10^-3

So, [OH-] = x = 7.218*10^-3 M

use:

pOH = -log [OH-]

= -log (7.218*10^-3)

= 2.1416

use:

PH = 14 - pOH

= 14 - 2.1416

= 11.8584

Answer: 11.86