Answer to Question #145495 in General Chemistry for Abraham Alfeuss

Question

Hydrazine N2H4 can interact with water in two steps
N2H4 (aq) + H2O (l) ↔ N2H5+ (aq) + OH- (aq) Kb1 = 8.5 x 10-7
N2H5+ (aq) + H2O (l) ↔ N2H62+ (aq) + OH- (aq) Kb2 = 8.9 x 10-16
a) What is the concentration of OH- ; N2H5+ ; and N2H62+ in a 0.010 M solution of hydrazine?
b) What is the pH of the 0.010 M solution of hydrazine?

Answer 1

PH = Pkw+(0.5 Pkb )+ (0.5logc)

PH = 14 + 0.5× (-log 8.5×10^-7) + 0.5 log (0.01)

PH = 14 - 3.035 - 1

PH = 9.65

POH = 14 - 9.65 = 4.035

POH = -log (OH^-)

4.035 = -log OH

OH^- = 9.23 ×10^-5 M

Kb₁= [N2H5][OH-]/[N2H4]

N2H5 = (8.5 ×10^-7 × 0.01)/ (9.23×10^-5)

= 9.21 × 10^-5M

Kb₂ = [N2H6][OH] / N2H5

N2H6 = (8.9 ×10^-16×9.21 ×10^-5 )/( 9.23 × 10^-5 )

= 8.88 ×10^-16M

b. PH = 9.65