Question

Hydrazine (H₃N⁺–⁺NH₃) is a diprotic base (the basic protons are bolded) with the following pK_a’s:

pK_a1 = –0.99

pK_a2 = 8.02

1. Write the two reactions that correspond to the two K_a’s.

2. Write the reaction that corresponds to the K_b for the conjugate base of the monoprotonated hydrazine (H₃N⁺–NH₂). Calculate K_b for this reaction.

3. Calculate the pH of a 0.500 M aqueous solution of hydrazine assuming that the only hydrazine species in solution are the unprotonated and the monoprotonated forms and the concentration of the diprotonated cation is negligible (=0).

Answer 1

Hydrazine (N2H4) is a weak base which can accept two protons as-

i- N2H4 + H2O ------------> N2H5+ + OH- pKa1 = –0.99

Now here the reaction is written as reaction of base, but the pKa is for reaction of acid i.e

N2H5+ + H2O ------------> N2H4 + H₃O⁺ Ka1 = 10^-pKa1 = 10^-.99 = 0.102

That means pKb1 = 14 - pKa1

= 14 - ( –0.99)

= 14 + 0.99

= 14.99

So here Kb1 = 10^-pKb1 = 10^-14.99 = 1.023

ii- N₂H₅⁺ + H2O ------------> N₂H₆⁺² + + pKa2 = 8.02

Now here the reaction is written as reaction of base, but the pKa is for reaction of acid i.e

N₂H₆⁺² + H2O ------------> N₂H₅⁺ + H₃O⁺ Ka2 = 10^-pKa2 = 10^-8.02 = 9.55 * 10⁹  

That means pKb2 = 14 - pKa2

= 14 - ( 8.02)

= 5.98‬

So here Kb2 = 10^-pKb2 = 10^5.98 = 9.55 * 10⁵  

3-

Now given 0.500 M aqueous solution of hydrazine assuming that the only hydrazine species in solution are the unprotonated and the monoprotonated forms

i.e N2H4 + H2O ------------> N2H5+ + OH-

Here the reaction is written in base reaction form

So if we take initially only  0.500 M of N2H4 is present, then

reaction	N2H4	N2H5+	OH-
Initial	0.500 M	0	0
Change	-x	+x	+x
Equilibrium	0.500 -x	x	x

Here Kb = [N2H5+]* [ OH-] / [N2H4]

Then putting the values-

Kb = [N2H5+]* [ OH-] / [N2H4]

1.023 = [x]* [ x] / [0.500 -x]

1.023 * [0.500 -x] = [x]* [ x]

0.5115 - 1.023x = x²  

x² + 1.023x - 0.5115 = 0

Solving this

x = 0.367 M

So

[OH- ] = x = 0.367 M

So

pOH = -log[OH- ] = -log[0.367 ] = 0.435

So

pH = 14 - pOH

= 14 - 0.435

= 13.565