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Topic: Buffer ( AGAIN )  (Read 14573 times)

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Offline uzi4u2

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Buffer ( AGAIN )
« on: August 19, 2006, 11:10:56 AM »
hello again my angelic helpers  ;)
small american question to solve :

Which of the following mixtures results a PH=7.2 buffer ?
( phosphoric acid pKa1=2.1  pKa2=7.2,  pKa3= 12.3 )

1)   100ml 1M Na2HPO4  +  100ml  1M NaH2PO4
2)   100ml 1M H3PO4     +  100ml  1M  NaOH
3)   100ml 1M Na2HPO4 +  100ml  1M Na3PO4
4)   100ml 1M H3PO4    +   100ml  1M NaH2PO4

the answer should be 1 but what formula do i use ?  i mean do i do it like this :

100ml * 1   ( only 1 H    for Na2HPO4 )       100
100ml * 2   (Two H  for NaH2PO4 )            200

then    200-100= 100      which means we will have 100 in a 200 solution
now what ?

Offline Albert

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Re: Buffer ( AGAIN )
« Reply #1 on: August 19, 2006, 11:22:55 AM »
I would say you don't need any forumula, since pH = pKa2 (you noticed that, didn't you?). However, if you like formulae, you can you H&H:  pH = pKa2 + log [base]/[acid] = pka2 + log [Na2HPO4]/[NaH2PO4].

Now, since [Na2HPO4] = [NaH2PO4], you have that pH = pKa2 = 7.2

Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #2 on: August 19, 2006, 11:40:05 AM »
ummm but it dosent come out :/
i am doing somethig wrong here
so you said  doing    [Na2HPO4]/[NaH2PO4]    which looks like this :
 
         100*1*1  /  100*1*2   =  0.5

so how do i go from here ?  i dunno maybe its too easy and i am making it too complicated :\

Offline Albert

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Re: Buffer ( AGAIN )
« Reply #3 on: August 19, 2006, 11:52:09 AM »
100ml * 1   ( only 1 H    for Na2HPO4 )       100
100ml * 2   (Two H  for NaH2PO4 )            200

What is this?

Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #4 on: August 19, 2006, 12:00:13 PM »
calculating thier concentrations ? i guess  :-\

Offline Albert

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Re: Buffer ( AGAIN )
« Reply #5 on: August 19, 2006, 12:11:36 PM »
You mean, calculating the moles, don't you?  ;)

Well, in order to do it, all you have to do is multiplying the molarity by the volume. Hence: 0.1 moles of sodium dihydrogen phosphate (your acid) react with 0.1 moles of disodium hydrogen phosphate (your base).

What you're trying to do, in my opinion, looks like considering this reaction as H2SO4 + 2NaOH, or something like that. Well, that's wrong.  ;)

For example, how would you calculate the pH of number 2?   ???

Quote
100ml 1M H3PO4     +  100ml  1M  NaOH

 :)

Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #6 on: August 19, 2006, 01:31:49 PM »
thx for the help dude :)

Offline Albert

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Re: Buffer ( AGAIN )
« Reply #7 on: August 20, 2006, 05:00:42 AM »
You're welcome, but I was really asking you how you'd calculate the pH of option #2, so that I could clarify your doubts.

Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #8 on: August 20, 2006, 11:26:12 AM »
okay you did point out that the calculation i was trying to do is a reaction calculation good .

so your saying i gotte multiply the molarity by the volume okay so lets take #2 as an example :

100ml 1M H3PO4     +  100ml  1M  NaOH
  so
                    NaOH/H3PO4             nothing is chanceled which means the % isnt 1:1 and then the
                                                   Ph will not be the same as the Pkb     am i right ?

 
   

Offline Borek

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Re: Buffer ( AGAIN )
« Reply #9 on: August 20, 2006, 11:41:20 AM »
100ml 1M H3PO4     +  100ml  1M  NaOH

Acid+base - they react. You have to find out what will be in the solution once the reaction ends. Very simple stoichiometry in this case.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #10 on: August 20, 2006, 12:48:39 PM »
okay here are 2 other buffer questions :  using numbrs and not theory :

How many ml of 1 N HCL should be added to 1 L 0.1M isoelectric Arginine solution to obtain a buffer with p.h=9.0 
pKa values of Arginine are 2.2 9.0 and 12.5
a]150
b]100
c]5
d]10
e] 50
                       how do i solve it ? i mean i cant use the regular equation for it can i ?

10 ml 0.1M Glutamate solution ( starting from fully protonated form ) is tirtated with 0.1 N NaOH.
at which of the NaOH additions do you expect buffer formation ?

a] 5ml
b]10ml
c]15ml
d]20
e]25
f]30
                                         also no idea :/   i know i gotte find the Meq somehow

Offline Borek

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Re: Buffer ( AGAIN )
« Reply #11 on: August 20, 2006, 02:14:29 PM »
Write reaction equations for both questions (and for every neutralization step).
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Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #12 on: August 20, 2006, 03:01:30 PM »
and how do i do that ? can you liek gimme an example ?

Offline Borek

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Re: Buffer ( AGAIN )
« Reply #13 on: August 20, 2006, 03:13:20 PM »
and how do i do that ? can you liek gimme an example ?

You really don't know how to write stepwise neutralization reactions? Something like

H2A + OH- -> HA- + H2O
HA- + OH- -> A2- + H2O

I hope it was just a bad joke  :-\
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Offline uzi4u2

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Re: Buffer ( AGAIN )
« Reply #14 on: August 20, 2006, 03:27:06 PM »
OOOOOOOOOOOOOOOOO
okay so after i write the equations , what do i do with them ? i mean how will they help me figure out the quantity i need ?

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