Question

Calculate the percent ionization in each of the following solutions. a.) 0.10 M NH₃ b.) 0.010 M NH₃

Answer 1

Part a

for NH3 pKb=4.75

.. .. .. .. .. .. NH3 + H2O <=> NH4+ +OH-
Initial .. .. .. . C
React .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. x .. .. .. x
At Equil .. .. C-x .. .. .. .. .. .. .. .x .. .. .. x

In such cases [H2O] is considered constant and thus it is not included in the expression of equilibrium constants and the ICE table

Kb= [NH4+][OH-]/[NH3] = x^2/(C-x) (1)

If we call the degree of dissociation a, then
a =mole/L that dissociated / initial mole/L = x/C =>
x=aC (2)

Substitute x in (1) using (2)

Kb= a^2C^2/(C-aC)= a^2C/(1-a) =>
C*a^2 + Kb*a- Kb=0
C=0.1
Kb=10^-pKb= 10^-4.75 = 1.78*10^-5
so the equation is
0.1a^2 +1.78*10^-5 a -1.78*10^-5 =0
so a=0.013= 1.3 %

Part B) Replace 0.235M to  0.010M so you can calculate for 0.010M

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

Kb = 1.8 x 10^-5 = x^2/ 0.235-x

x = 0.00206

% ionization = 0.00206 x 100/ 0.235 = 0.875