Question

What is the pH of a 0.225 Mammonia solution? Express your answer numerically to two decimal places. View Available Hint(s) IVO ALVA O ? pH = Submit Request Answer Part B What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units. View Available Hint(s) TH HÀ • • • Ea ? % ionization = Value Units

Answer 1

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 %.