If Kb for NX3 is 8.5×10−6, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Answer 1

NX₃ + H₂O ==> HNX₃⁺ + OH^-

Kb = 8.5x10^-6 = [HNX₃⁺]{OH^-] / [NX₃]

8.5x10^-6 = (x)(x) / 0.325 - x (and assuming x is small we can ignore it in the denominator)

x² = 2.76x10^-6

x = 1.66x10^-3 = [OH^-] = [HNX₃⁺] (the above assumption is valid as seen by % ionization below)

% ionization = 1.66x10-3 M / 0.325 M (x100%) = 0.51% ionization