Question

If kb for nx3 is 1.5×10−6, what is the percent ionization of a 0.325 m aqueous solution of nx3?

Answer 1

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
c(NX₃) = 0,325 M.
Kb = 1,5·10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0,325 M - x.
Kb = [NX₃H⁺] · [OH⁻] / [NX₃]. 
1,5·10⁻⁶ = x² / (0,325 M - x).
x = 0,0007 M.
percent of ionization:
α = 0,0007 M ÷ 0,325 M · 100% = 0,215%.