Question

1. Calculate the pH of a 0.417 M aqueous solution of isoquinoline (C₉H₇N, K_b = 2.5×10^-9).

2. Calculate the pH of a 0.0351 M aqueous solution of methylamine (CH₃NH₂, K_b = 4.2×10^-4).

3. Calculate the pH of a 0.0432 M aqueous solution of the weak base diethylamine((C₂H₅)₂NH, K_b = 6.90×10^-4).

Answer 1

1)

C9H7N dissociates as:

C9H7N +H2O -----> C9H7NH+ + OH-

0.289 0 0

0.289-x x x

Kb = [C9H7NH+][OH-]/[C9H7N]

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 ((2.5*10^-9)*0.289) = 2.688*10^-5

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

so, x = 2.688*10^-5 M

So, [OH-] = x = 2.688*10^-5 M

use:

pOH = -log [OH-]

= -log (2.688*10^-5)

= 4.5706

use:

PH = 14 - pOH

= 14 - 4.5706

= 9.4294

Answer: 9.43

2)

CH3NH2 dissociates as:

CH3NH2 +H2O -----> CH3NH3+ + OH-

3.51*10^-2 0 0

3.51*10^-2-x x x

Kb = [CH3NH3+][OH-]/[CH3NH2]

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 ((4.2*10^-4)*3.51*10^-2) = 3.84*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)

4.2*10^-4 = x^2/(3.51*10^-2-x)

1.474*10^-5 - 4.2*10^-4 *x = x^2

x^2 + 4.2*10^-4 *x-1.474*10^-5 = 0

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

a = 1

b = 4.2*10^-4

c = -1.474*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 = 5.914*10^-5

roots are :

x = 3.635*10^-3 and x = -4.055*10^-3

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

x = 3.635*10^-3

So, [OH-] = x = 3.635*10^-3 M

use:

pOH = -log [OH-]

= -log (3.635*10^-3)

= 2.4395

use:

PH = 14 - pOH

= 14 - 2.4395

= 11.5605

Answer: 11.56

3)

(C2H5)2NH dissociates as:

(C2H5)2NH +H2O -----> (C2H5)2NH2+ + OH-

4.32*10^-2 0 0

4.32*10^-2-x x x

Kb = [(C2H5)2NH2+][OH-]/[(C2H5)2NH]

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 ((6.9*10^-4)*4.32*10^-2) = 5.46*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)

6.9*10^-4 = x^2/(4.32*10^-2-x)

2.981*10^-5 - 6.9*10^-4 *x = x^2

x^2 + 6.9*10^-4 *x-2.981*10^-5 = 0

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

a = 1

b = 6.9*10^-4

c = -2.981*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 = 1.197*10^-4

roots are :

x = 5.126*10^-3 and x = -5.816*10^-3

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

x = 5.126*10^-3

So, [OH-] = x = 5.126*10^-3 M

use:

pOH = -log [OH-]

= -log (5.126*10^-3)

= 2.2903

use:

PH = 14 - pOH

= 14 - 2.2903

= 11.7097

Answer: 11.71