Question

Determine the pH of a solution that is 0.034 M C2H8N2 where Kb(C2H8N2) = 8.3 * 10−5?

Answer 1

C2H8N2 dissociates as:

C2H8N2 +H2O -----> C2H8N2H+ + OH-

3.4*10^-2 0 0

3.4*10^-2-x x x

Kb = [C2H8N2H+][OH-]/[C2H8N2]

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 ((8.3*10^-5)*3.4*10^-2) = 1.68*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)

8.3*10^-5 = x^2/(3.4*10^-2-x)

2.822*10^-6 - 8.3*10^-5 *x = x^2

x^2 + 8.3*10^-5 *x-2.822*10^-6 = 0

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

a = 1

b = 8.3*10^-5

c = -2.822*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 = 1.129*10^-5

roots are :

x = 1.639*10^-3 and x = -1.722*10^-3

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

x = 1.639*10^-3

so.[OH-] = x = 1.639*10^-3 M

use:

pOH = -log [OH-]

= -log (1.639*10^-3)

= 2.79

use:

PH = 14 - pOH

= 14 - 2.79

= 11.21

Answer: 11.21