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Topic: Can someone please help (phys chem)  (Read 14763 times)

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Offline Lisa_G

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Can someone please help (phys chem)
« on: March 04, 2008, 05:32:00 PM »
The weak base hydrazine. has the chemical formula NH2 NH2

(pKb1 = 5.77; pKb2 = 15.05)

Write balanced equations showing the first and second base dissociations, relating to Kb1 and Kb2 in water.

Any answer would be appreciated
Regards, Lisa

« Last Edit: March 04, 2008, 07:31:05 PM by Lisa_G »

Offline agrobert

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Re: Can someone please help (phys chem)
« Reply #1 on: March 04, 2008, 05:36:19 PM »
Attempt to write your first equation.  If you don't know how to incorporate the pkb then just show the product side of hydrazine in water.  What is the pka / pkb of water?
In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur

Offline Borek

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Re: Can someone please help (phys chem)
« Reply #2 on: March 04, 2008, 05:37:50 PM »
Please read forum rules.

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

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Re: Can someone please help (phys chem)
« Reply #3 on: March 04, 2008, 07:21:34 PM »
All ive been given is the above information and thats it :(

Offline agrobert

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Re: Can someone please help (phys chem)
« Reply #4 on: March 04, 2008, 07:23:04 PM »
Please attempt the problem.

Hint:

The first product is NH2NH3+
In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur

Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #5 on: March 04, 2008, 07:51:50 PM »
H2O + NH2NH2 ----> NH2NH3 + OH-

thats what i got, but i dont understand the Kb1 and Kb2 bits

Offline agrobert

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Re: Can someone please help (phys chem)
« Reply #6 on: March 04, 2008, 07:59:01 PM »
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 kb1.  What gets protonated next?  What is kb2?

Do you know how to convert between Kb and pkb?
In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur

Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #7 on: March 05, 2008, 06:36:47 PM »
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 :(

Offline Borek

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Re: Can someone please help (phys chem)
« Reply #8 on: March 06, 2008, 03:01:26 AM »
You are very close - there is just a small mistake in your second equation. What should be your reactant there (apart from water)?
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Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #9 on: March 06, 2008, 04:43:43 PM »
errrm instead water would it be H3O+
dont see anything else fitting in to be honest

Offline agrobert

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Re: Can someone please help (phys chem)
« Reply #10 on: March 06, 2008, 05:14:34 PM »
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 <--(Kb1)--> -OH + NH2NH3+
(2)   H2O + NH2NH3+ <--(Kb2)--> -OH + +NH3NH3+
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Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #11 on: March 06, 2008, 05:27:16 PM »
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.

Offline Borek

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Re: Can someone please help (phys chem)
« Reply #12 on: March 07, 2008, 03:45:41 AM »
see im totally useless :(

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

Quote
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?

Quote
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.
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Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #13 on: March 09, 2008, 12:16:20 PM »
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

Offline Lisa_G

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Re: Can someone please help (phys chem)
« Reply #14 on: March 09, 2008, 12:54:15 PM »
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

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