Respuesta :

Answer:

% α = 0.1577 %

Explanation:

  • H2NNH2(aq) + H2O(l) ↔ H2NNH3+(aq) + OH-(aq)
  • 0.5216 - X                               X                      X

∴ 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

  • % α = (X/molality)*100......percent ionization

⇒ 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 %

pstnonsonjoku

The percent dissociation of hydrazine is  0.16 %.

The dissociation equation is;

             H₂NNH₂    +  H2O ⇄     H3O^+   +   HNNH₂^-

I            0.520                                 0            0

C            -x                                      +x           +x

E         0.520 -x                               x              x

Where, Kb = 1.3 × 10⁻⁶

Kb = [x] [x]/[ 1.3 × 10⁻⁶]

1.3 × 10⁻⁶ = x^2/[ 0.520 -x]

1.3 × 10⁻⁶ [ 0.520 -x] = x^2

6.76 × 10⁻7 -  1.3 × 10⁻⁶x = x^2

x^2 + 1.3 × 10^-6x - 6.76 × 10^-7 = 0

x = 0.00082 M

Hence percent dissociation = [HNNH₂^-]/[H₂NNH₂ ] × 100/1

percent dissociation = 0.00082 M/ 0.520 M × 100/1

= 0.16 %

Learn more: https://brainly.com/question/11897796

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