Answer:
%Ionization = 1.63%
Explanation:
Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.
H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶
So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.
HONHNHOH => HONHNH⁺ + OH⁻
C(i) => 0.490M 0M 0M
ΔC => -x +x +x
C(eq) => 0.490 - x x x
≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100
Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]
1.3 x 10⁻⁶ = x²/0.490
=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴
=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%
