Question

The salt formed by the reaction of the weak base aniline, C6H5NH2, with the strong acid nitric acid is anilinium nitrate, C6H5NH3NO3. What is the hydronium ion concentration of a 0.158 M solution of anilinium nitrate at 25∘C given that the value of Kb for aniline is 4.300×10−10?

• Report your answer in scientific notation.
• Your answer should have three significant figures.

Answer 1

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+

0.158 0 0

0.158-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)*0.158) = 1.917*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/(0.158-x)

3.674*10^-6 - 2.326*10^-5 *x = x^2

x^2 + 2.326*10^-5 *x-3.674*10^-6 = 0

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

a = 1

b = 2.326*10^-5

c = -3.674*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 = 1.47*10^-5

roots are :

x = 1.905*10^-3 and x = -1.929*10^-3

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

x = 1.905*10^-3

So, [H+] = x = 1.905*10^-3 M

use:

pH = -log [H+]

= -log (1.905*10^-3)

= 2.72

Answer: 2.72