Question

hydrazine N2H4 can interact with water in two steps,
N2H4(aq)+ H2O(l)=N2H5(aq)+OH(aq) Kb1=8.5x10^-7,
N2H5(aq)+ H2O(l)=N2H6(aq)+OH(aq) Kb2=8.9x10^-16 ,
(a) what is the concentration of OH-; N2H5+ and N2H6^2+ in a 0.010M solution of hydrazine?
(b) what is the pH of the 0.010M solution of hydrazine?

Answer 1

PH = Pkw+1/2 Pkb + 1/2logc

PH = 14 + 1/2 × (-log 8.5×10^-7) + 1/2 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

PH = 9.65 as solved earlier