Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: uzi4u2 on August 19, 2006, 11:10:56 AM
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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 ?
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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
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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 :\
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100ml * 1 ( only 1 H for Na2HPO4 ) 100
100ml * 2 (Two H for NaH2PO4 ) 200
What is this?
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calculating thier concentrations ? i guess :-\
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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? ???
100ml 1M H3PO4 + 100ml 1M NaOH
:)
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thx for the help dude :)
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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.
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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 ?
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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.
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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
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Write reaction equations for both questions (and for every neutralization step).
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and how do i do that ? can you liek gimme an example ?
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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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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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Write them first.
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okay so :
HCL + C6H14N4O2 ---> h2o + ( stuck here )
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Sorry, no spoon feeding.
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okay i agree with you , then why not just give me one of them and then i can do the other based on the first ;D
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please please pleaseeeee i realy need to answer this question , got achemistry final and this calculation or somethng like it will surely be in it :(