Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3NH2(aq) solution. (Kb for CH3NH2 = 4.4 × 10-4)

 The percent ionization of in a 0.050 M solution is 8.9 %

Explanation:

 cM                            0             0

So dissociation constant will be:

Give c= concentration = 0.050 M and = degree of ionisation = ?

Putting in the values we get:

percent ionisation =

i think its 100

i hope this !