[Lisa_G]

The weak base hydrazine. has the chemical formula NH₂ NH₂

(pK_b1 = 5.77; pK_b2 = 15.05)

Write balanced equations showing the first and second base dissociations, relating to K_b1 and K_b2 in water.

Any answer would be appreciated
Regards, Lisa

[agrobert]

Attempt to write your first equation.  If you don't know how to incorporate the pk_b then just show the product side of hydrazine in water.  What is the pk_a / pk_b of water?

[Borek]

Please read forum rules.

Hint: reactions will be very similar to that of ammonia.

[Lisa_G]

All ive been given is the above information and thats it

[agrobert]

Please attempt the problem.

Hint:

The first product is NH₂NH₃⁺

[Lisa_G]

H₂O + NH₂NH₂ ----> NH₂NH₃ + OH-

thats what i got, but i dont understand the K_b1 and K_b2 bits

[agrobert]

Ok the charge in your first equation is not balanced.

http://en.wikipedia.org/wiki/Base_dissociation_constant#Bases

You have protonated one basic end of your hydrazine.  That gives the protonated nitrogen a + formal charge (balancing your equation).  The constant that determines the rate at which this newly obtained proton is dissociated is k_b1.  What gets protonated next?  What is k_b2?

Do you know how to convert between K_b and pk_b?

[Lisa_G]

H₂O + NH₂NH₂ ----> NH₂NH₃⁺ + OH^-

yeh sorry i missed the + sign before

what gets protonated next i think would be the other NH2, giving me

H₂O + NH₂NH₂ ----> NH₃⁺NH₃⁺ + OH^-

i dont know if thats right,but the i got an unbalanced equation, what would happen to the  OH-. im completely useless at this

[Borek]

You are very close - there is just a small mistake in your second equation. What should be your reactant there (apart from water)?

[Lisa_G]

errrm instead water would it be H₃O⁺
dont see anything else fitting in to be honest

[agrobert]

errrm instead water would it be H₃O⁺
dont see anything else fitting in to be honest

No.  You have excess H₂O in a basic environment.  Your equations should be

(1)   H₂O + NH₂NH₂ <--(K_b1)--> ^-OH + NH₂NH₃⁺
(2)   H₂O + NH₂NH₃⁺ <--(K_b2)--> ^-OH + ⁺NH₃NH₃⁺

[Lisa_G]

see im totally useless thanks for helping soooo much

Anyway when [NH₂-NH₂]e = [NH₃⁺NH2]e (so would the equilibrium concentrations be exactly the same) i.e. pH is 5.77.

And when [NH₃⁺-NH₂]e = [NH₃⁺-NH₃⁺]e then the pH of the solution is 15.05.

[Borek]

see im totally useless

You are not. You were close. You just needed a small push.

Anyway when [NH₂-NH₂]e = [NH₃⁺NH2]e (so would the equilibrium concentrations be exactly the same) i.e. pH is 5.77.

Not sure if I like your notation. And once again - you are close, but you need a push. This is base, and you are given pKb, not pKa. 5.77 is slightly acidic - not a thing that sounds logical in the case of base solution. Think it over - have you found pH or pOH?

And when [NH₃⁺-NH₂]e = [NH₃⁺-NH₃⁺]e then the pH of the solution is 15.05.

You have used the same logic that you used in the previous question - so you are wrong in exactly the same way. However, there is more - this logic in water doesn't extend so far. Water is much stronger base and it is present in excess. Don't bother with pH of this solution if you don't have to, just ignore it for now.

[Lisa_G]

i've been working on this and have done some reading on it, for kb 1 and 2 i got the following

NH₂NH₂ + H₂O <--> NH₃⁺NH₂ + OH-

NH₃⁺NH2 + H2O <--> NH₃⁺NH₃⁺ + OH-

and for the overall reaction i got

2H₂O + NH2-NH2  <----> NH₃⁺NH₃⁺ + 2OH-

can you guide me on how to calculate the dissociation constant kb1 & kb2, the conccentration of the base is 0.2M

[Lisa_G]

since my last post ive been trying to work out the pH for kb1
what i got is

10^-pkb1 = kb1
10 ^{– 5.77} = kb1 = 1.70 x10 ^-6

1.70 x10 ^-6 = [ OH-]²
                     0.02

OH- = √ 1.70 x10 ^-6  X 0.02
= 1.84 x10 ^-4

pOH = - log [OH-]
        = - log (1.84 x10 ^-4)
        = 3.74

pH = 14 – pOH
      = 14 – 3.74
pH = 10.26

dont know if thats right but can someone correct me if i've gone wrong any where