Question

Calculate the pH of a 0.0760 M anilinium chloride solution. A list of ionization constants can be found here.

Calculate the concentrations of the conjugate acid and base at equilibrium. The conjugate acid and conjugate base are represented as HA and A–, respectively.

Answer 1

Kb of aniline = 4.3*10^-10

Aniline is C6H5NH2

Anilinium ion is C6H5NH3+

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/4.3*10^-10

Ka = 2.326*10^-5

C6H5NH3+ + H2O -----> C6H5NH2 + H+

7.6*10^-2 0 0

7.6*10^-2-x x x

Ka = [H+][C6H5NH2]/[C6H5NH3+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.326*10^-5)*7.6*10^-2) = 1.329*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

2.326*10^-5 = x^2/(7.6*10^-2-x)

1.767*10^-6 - 2.326*10^-5 *x = x^2

x^2 + 2.326*10^-5 *x-1.767*10^-6 = 0

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

a = 1

b = 2.326*10^-5

c = -1.767*10^-6

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 = 7.07*10^-6

roots are :

x = 1.318*10^-3 and x = -1.341*10^-3

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

x = 1.318*10^-3

So, [H+] = x = 1.318*10^-3 M

Acid is C6H5NH3+

[C6H5NH3+] = 0.0760 - x

= 0.0760 - 1.318*10^-3

= 0.0747 M

conjugate base is C6H5NH2

[C6H5NH2] = x = 1.318*10^-3 M

[acid] = 0.0747 M

[conjugate base] = 1.32*10^-3 M