Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Lisa_G on March 04, 2008, 05:32:00 PM
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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
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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?
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Please read forum rules (http://www.chemicalforums.com/index.php?page=forumrules).
Hint: reactions will be very similar to that of ammonia.
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All ive been given is the above information and thats it :(
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Please attempt the problem.
Hint:
The first product is NH2NH3+
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H2O + NH2NH2 ----> NH2NH3 + OH-
thats what i got, but i dont understand the Kb1 and Kb2 bits
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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?
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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 :(
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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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errrm instead water would it be H3O+
dont see anything else fitting in to be honest
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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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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.
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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.
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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
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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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pH = 10.26
That's OK. Could be you have accidentally used correct equation, but 10.26 it is.
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thanks for that, i calculated for kb2 as well and i got pH 5.63 using the same method, is it suppose to be below or above 7
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You can't calculate effects of the second dissociation step the same way. However, second dissociation constant is so low, you can safely ignore it.
More on pH calculation here:
http://www.chembuddy.com/?left=pH-calculation&right=toc
and specific information about calculation of multiprotic acid pH:
http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified
(note that for base you just replace Ka with Kb and H+ with OH-).
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thanks for that information, very helpful
i got this question
a buffer solution contains equal volumes of H3PO4 (0.5M) and its conj base H2PO4- (0.45M)
Acid dissociation constant Ka = 7.5 x10 -3
What is the pH
Because the volumes are equal i used the henderson hasslebach equation and got
pH = (-log 7.5 x10-3) + log 0.45
0.5 pH = 2.08
is that right ?
and the 2nd question i got is
10 cm3 of 0.05M H3O+ is added to 1dm3 of the solution above. what is the new pH ?
i got
H3O+ moles = 10
1000 x 0.05 = 5x10-4
AFter addition of H3O+ (in 1dm3) H3PO4 = 0.5 - 5x10-4 = 0.50 moles
[H3PO4] = 1000
1010 x 0.50 = 0.50M
H2PO4- = 0.45 + 5x10-4 = 0.45 moles
[H2PO4-] = 1000
1010 x 0.45 = 0.46M
pH = (- log 7.5 x10-3) + log 0.46
0.50
pH = 2.08
is the method i used correct
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2.08 twice is OK, but your calculations are hard to follow. Better write it like pH=2.12-log(0.45/0.5), it is much more obvious what you are doing.
0.5 - 5x10-4 = 0.50 is not true. You should never round down intermediate results.
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yeh i listened to what you said, i got 2.08 and 2.07, thanks for all your help, i really appreciate it