Question

Determine the pH of a 0.125 Molar NH3 solution

Answer 1

Kb for NH3 = 1.8*10^-5

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.125 0 0

0.125-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-5)*0.125) = 1.5*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)

1.8*10^-5 = x^2/(0.125-x)

2.25*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-2.25*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -2.25*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 = 9*10^-6

roots are :

x = 1.491*10^-3 and x = -1.509*10^-3

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

x = 1.491*10^-3

so.[OH-] = x = 1.491*10^-3 M

use:

pOH = -log [OH-]

= -log (1.491*10^-3)

= 2.83

use:

PH = 14 - pOH

= 14 - 2.83

= 11.17

Answer: 11.17