Question

Calculate the [OH-] and pH of a 0.01 M solution of ammonia (Kb= 1.8 x 10^-5)

Answer 1

NH3 dissociates as:

NH3 +H2O     ----->     NH4+   +   OH-
1*10^-2                   0         0
1*10^-2-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)*1*10^-2) = 4.243*10^-4

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/(1*10^-2-x)
1.8*10^-7 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-1.8*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -1.8*10^-7

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 = 7.203*10^-7

roots are :
x = 4.154*10^-4 and x = -4.334*10^-4

since x can't be negative, the possible value of x is
x = 4.154*10^-4

so,
[OH-] = x = 4.154*10^-4 M


use:
pOH = -log [OH-]
= -log (4.154*10^-4)
= 3.38


use:
PH = 14 - pOH
= 14 - 3.3816
= 10.62

Answer:
[OH-] = 4.154*10^-4 M
pH = 10.62