An aqueous solution contains 0.050 m of methylamine, with kb = 4.4 x 10-4. what is the percent ionization and the ph of the solution?

The base dissociation constant or Kb is a value used to measure the strength of a specific base in solution. To determine the percent ionization of the substance we make use of the Kb given. Methylamine or CH3NH2 when in solution would form ions:

CH3NH2 + H2O < = >  CH3NH3+ + OH- 

Kb is expressed as follows:

Kb = [OH-] [CH3NH3+] / [CH3NH2]

Where the terms represents the concentrations of the acid and the ions. 

By the ICE table, we can calculate the equilibrium concentrations,
     CH3NH2    CH3NH3+             OH- 
I      0.05            0                         0
C      -x               +x                      +x
 
E  0.05-x             x                       x

Kb = [OH-] [CH3NH3+] / [CH3NH2] = 4.4 x 10^-4

Solving for x,

x = [OH-] = 0.00448 M 

percent ionization = 0.00448 / 0.05 x 100 = 8.96%

pH = 14 + log 0.00448 = 11.65

explanation:

i think option no d is the correct answer

14 days is the half-life of phosphorus-32