Question

Calculate the pH of a 0.0438 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and the equilibrium concentrations of the weak base and its conjugate acid.

Calculate the pH of a 0.289 M aqueous solution of quinoline (C₉H₇N, K_b = 6.3×10^-10) and the equilibrium concentrations of the weak base and its conjugate acid.

Answer 1

1)

(CH3)2NH dissociates as:

(CH3)2NH +H2O -----> (CH3)2NH2+ + OH-

4.38*10^-2 0 0

4.38*10^-2-x x x

Kb = [(CH3)2NH2+][OH-]/[(CH3)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 ((5.9*10^-4)*4.38*10^-2) = 5.084*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)

5.9*10^-4 = x^2/(4.38*10^-2-x)

2.584*10^-5 - 5.9*10^-4 *x = x^2

x^2 + 5.9*10^-4 *x-2.584*10^-5 = 0

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

a = 1

b = 5.9*10^-4

c = -2.584*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 = 1.037*10^-4

roots are :

x = 4.797*10^-3 and x = -5.387*10^-3

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

x = 4.797*10^-3

So, [OH-] = x = 4.797*10^-3 M

[(CH3)2NH] = 0.0438 - x

= 0.0438 - 4.797*10^-3

= 0.0390 M

[(CH3)2NH2+] = x = 4.797*10^-3 M

use:

pOH = -log [OH-]

= -log (4.797*10^-3)

= 2.319

use:

PH = 14 - pOH

= 14 - 2.319

= 11.681

PH = 11.68

[(CH3)2NH] = 0.0390 M

[(CH3)2NH2+] = 4.80*10^-3 M

2)

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 ((6.3*10^-10)*0.289) = 1.349*10^-5

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

so, x = 1.349*10^-5 M

[C9H7N] = 0.289 - x

= 0.289 - 1.349*10^-5

= 0.289 M

[C9H7NH+] = x

= 1.35*10^-5 M

So, [OH-] = x = 1.349*10^-5 M

use:

pOH = -log [OH-]

= -log (1.349*10^-5)

= 4.8699

use:

PH = 14 - pOH

= 14 - 4.8699

= 9.1301

pH = 9.13

[C9H7N] = 0.289 M

[C9H7NH+] = 1.35*10^-5 M