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Topic: buffer  (Read 9222 times)

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Offline kapital

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buffer
« on: September 03, 2011, 06:28:42 AM »
How can you preprare  500 ml buffer from  0,600 M H3PO4 and 3,00 M NaOH.(pH=3)

pKa1=2,148

[H2PO4-]/[H3PO4]=Ka1/[H30+]=7,11

Out of there I get 293,3 ml NaOH and 206,61 ml of H3PO4, but thats wrong, why?

(There is not mistake that Ka2 and Ka3 of phosporic acid arent give, its simplifed)

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Offline sjb

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Re: buffer
« Reply #1 on: September 03, 2011, 09:04:59 AM »
Quote from: kapital on September 03, 2011, 06:28:42 AM
How can you preprare  500 ml buffer from  0,600 M H3PO4 and 3,00 M NaOH.(pH=3)

pKa1=2,148

[H2PO4-]/[H3PO4]=Ka1/[H30+]=7,11

Out of there I get 293,3 ml NaOH and 206,61 ml of H3PO4, but thats wrong, why?

(There is not mistake that Ka2 and Ka3 of phosporic acid arent give, its simplifed)



I have not done the maths here, but are you taking into account the concentrations of acid and base? Seems to me that a mixture as you describe will be heavily basic?
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Offline kapital

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Re: buffer
« Reply #2 on: September 03, 2011, 10:11:11 AM »
From that equation I get the base/acid ratio, than from eqution V(base)=500-V(acid) I get the results. ??
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Offline sjb

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Re: buffer
« Reply #3 on: September 03, 2011, 10:17:06 AM »
Quote from: kapital on September 03, 2011, 10:11:11 AM
From that equation I get the base/acid ratio, than from eqution V(base)=500-V(acid) I get the results. ??


Molar ratio, yes, not volume ratio.
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Offline kapital

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Re: buffer
« Reply #4 on: September 03, 2011, 04:39:17 PM »
Yes I know that. Can you please tell me how can I solve this exercise?

(What I am not completly sure, but does a reaction make any importance?

Since only first dissociation constant matter I write reactions as

H3PO4 + NaOH --> NaH2PO4 + H20)
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Offline Borek

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Re: buffer
« Reply #5 on: September 03, 2011, 05:32:39 PM »
Your reaction is correct.

You have a system of several equations. HH gives you molar ratio of the acid and conjugate base. You know sum of volumes must be 500 mL. These are two equations. Try to express final concentrations of acid and conjugate base using known concentrations, known final volume and two variables - VNaOH and VH3PO4. Solve.
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Offline kapital

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Re: buffer
« Reply #6 on: September 03, 2011, 05:43:40 PM »
I tried that, I get the same results. I tried everthing I can think off.  I dont know what is wrong.
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Offline Borek

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Re: buffer
« Reply #7 on: September 04, 2011, 04:09:21 AM »
Show your work then, and we will look for errors.
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Offline kapital

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Re: buffer
« Reply #8 on: September 04, 2011, 04:56:01 AM »
Ok. And I have another(2) one, the exercise is:

2. How many ml 0,1000 M HCl and how much  g Na-acetate dihydrate is needed for 250 ml buffer ,    pH 5,00 (5 C) and ionic strnght 0,100 M ?
pKw (5 C) = 14,734
pKa (5 C) = 4,77
Na=23

The amaunt of sodium acetate dihydrate is calculated corect, the amaunt of HCl is not.
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Offline kapital

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Re: buffer
« Reply #9 on: September 04, 2011, 04:58:36 AM »
2
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Offline Borek

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Re: buffer
« Reply #10 on: September 04, 2011, 06:28:40 AM »
When calculating concentration of phosphoric acid you have to take into account fact that part of it was neutralized.
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Offline kapital

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Re: buffer
« Reply #11 on: September 04, 2011, 08:02:23 AM »
How can I do that? And what abaut the second?
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Offline kapital

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Re: buffer
« Reply #12 on: September 04, 2011, 10:09:06 AM »
Can anabody please tell me how can I solve this two exercises, perhaps in some other way?
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Offline Borek

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Re: buffer
« Reply #13 on: September 04, 2011, 01:50:00 PM »
There is no other way. The way you tried you are assuming concentration of phosphoric acid is that of the acid that was put in the solution - but part of it was neutralized to H2PO4-, so you have to subtract neutralized amount from the amount that was put into the solution. This is all in the neutralization stoichiometry.
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Offline kapital

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Re: buffer
« Reply #14 on: September 04, 2011, 03:10:12 PM »
Can you show that please?
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