What is the percent ionization of H₂NNH₂ in a solution with a concentration of a 0.430 M? (Kb = 1.3 × 10⁻⁶)

Answer 1

H₂NNH₂ + H₂O ==> H₂NNH₃⁺ + OH^-

Kb = 1.3x10^-6 = [H₂NNH₃⁺][OH-] / [H₂NNH₂]

1.3x10^-6 = (x)(x) / 0.430

x² = 5.59x10^-7

x = 7.5x10^-4 = [H₂NNH₃⁺] = [OH-]

% ionization = 7.5x10^-4 / 0.430 (x100%)

% ionization = 0.17%