Question

A) What is the degree of ionization of ethylamine, C2H5NH2, in an 0.90 M aqueous solution? B) What is the percent ionization of ethylamine, C2H5NH2, in an 0.90 M aqueous solution?

Answer 1

6.4 x 10^-4 is the Kb of C2H5NH2

C2H5NH2 + H2O <=> C2H5NH3⁺ + OH-

initial 0.90 .......................... 0 .................... 0

final ( 0.90 - x ) ................... x ....................... x

Kb = [OH-] [C2H5NH3+] / [C2H5NH2]

Kb = [x] [x] / [0.90 - x]

6.4 x 10^-4 x 0.90 = x² {x<<<0.90 so x neglected }

x = 2.4x10^-2 M

degree of ionization of ethylamine, C2H5NH2 = 0.024

percent ionization of ethylamine C2H5NH2, = ( 0.024 M / 0.90 M ) *100 = 2.66 %