What is the percent ionization of H₂NNH₂ in a solution with a concentration of a 0.520 M? (Kb = 1.3 × 10⁻⁶)

% α = 0.1577 %

Explanation:

∴ δH2O = (997 Kg/m³)*(m³/1000 L) = 0.997 Kg/L

∴ C H2NNH2 = (0.520 mol/L)*(L/0.997 Kg) = 0.5216 mol/Kg

∴ Kb = [OH-][H2NNH3+] / [H2NNH2] = 1.3 E-6

⇒ Kb = (X)*(X) / (0.5216 - X) = X²/(0.5216 - X) = 1.3 E-6

⇒ X² + 1.3 E-6X - 6.78 E-7 = 0

⇒ X = 8.227 E-4

⇒ % α = (8.227 E-4 / 0.5216 )*100

⇒ % α = 0.1577 %

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;