Answer:
Part A:
Given [NH3]=0.225 M
Since ammonia is a weak base, the base dissociation constant, Kb=1.8 x 10-5.
The equilibrium reaction is
NH3 + H2O <------> NH4+ + OH-
Initial 0.225 0 0
Change -x +x +x
Equilibrium 0.225-x x x
Kb=[NH4+][OH-]/[NH3]
1.8 x 10-5=x2/(0.225-x)
x2+(1.8 x 10-5)x-4.05 x 10-6=0
After solving this quadratic equation
x=0.002.
Therefore [OH-]=0.002 M
Then pOH=-log[OH-]
pOH=-log(0.002)=2.698.
Since pH+pOH=14
pH=14-pOH=14-2.698
pH=11.3.
Part B:
% Ionization of ammonia=([NH4+]eq/[NH3]initial) x 100
Given [NH3]initial=0.225 M, and [NH4+]=x=0.002 M
% Ionization=(0.002M/0.225M) x 100
% Ionization=0.888 %.
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