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Topic: Polyprotic buffers  (Read 11177 times)

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Offline HaroTaro

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Polyprotic buffers
« on: February 04, 2009, 06:18:47 AM »
Trisodium arsenate has 3 pKa's:2.1, 6.9, 11.5

I want to prepare a 1L arsenic aicd/arsenate-based buffer w/ a total concentration of 0.2 M and pH 6.5

Given 1M trisodium arsenate and 5 M HCl.

How many moles of HCl should be added to prepare the buffer?

I was thinking that I could use the Henderson-Hassalbeck equation. Since we want a pH of 6.5, we should use the pKa for the 2nd ionization pKa = 2.1 b/c it's the closest to 6.5.
From that, I got [A-] = 28.57% and [HA] = 71.43%
Do I then multiply the precentage by 1M? Is it right so far? I'm not too sure.

Offline Borek

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Re: Polyprotic buffers
« Reply #1 on: February 04, 2009, 06:39:36 AM »
Not A- and HA, but H2A- and HA2-. You start with A3-, so you have to add enough acid to convert all A3- to HA2- first, then to convert part of the HA2- to H2A-.
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Offline HaroTaro

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Re: Polyprotic buffers
« Reply #2 on: February 04, 2009, 12:26:26 PM »
Can you give me some hints on how to do that? Do I use the H-H equation?

Offline aldoxime_amine

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Re: Polyprotic buffers
« Reply #3 on: February 04, 2009, 03:05:09 PM »
Hello,

Can someone verify and tell me the answer to this problem. I am getting a highly weird answer:

A recipe of 0.037 ml given salt solution (approx. 7.7 mg of the salt) in approx 1L of given HCl solution.


(maybe i am going wrong because: according to me, total concentration =0.2M implies
[H3A]+[H2A-]+[HA2-]+[A3-]=0.2M
all equilibrium concentrations.)

Offline Borek

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Re: Polyprotic buffers
« Reply #4 on: February 04, 2009, 05:16:53 PM »
[H3A]+[H2A-]+[HA2-]+[A3-]=0.2M

That's correct, but your result is wrong.

HaroTaro: yes, use HH equation for ions I have listed in previous post. Note, that concetrations of fully dissociated and undissociated form will be so low they can be safely neglected.
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Offline aldoxime_amine

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Re: Polyprotic buffers
« Reply #5 on: February 05, 2009, 02:50:29 AM »
I think i realized my mistake:

The answer is (edited by Borek) of salt solution, diluted to one litre.

Also, [HA2-]=0.057 M approx
and [H2A-]=0.143 M approx
[A3-] and [H3A] are tending to zero.
All equilibrium concentrations.
« Last Edit: February 05, 2009, 03:10:52 AM by Borek »

Offline Borek

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Re: Polyprotic buffers
« Reply #6 on: February 05, 2009, 03:10:32 AM »
Aldoxime, your answer was correct, but you have given straight answer to HaroTaro. That's against forum rules.
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Offline aldoxime_amine

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Re: Polyprotic buffers
« Reply #7 on: February 05, 2009, 03:23:51 AM »
Thanks.  ;D

But if we are not allowed to give an answer as such, then how can others who are trying to solve the problem be assured that they are doing it correctly?

Offline Borek

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Re: Polyprotic buffers
« Reply #8 on: February 05, 2009, 05:34:53 AM »
Once they answer, you can always tell them if they are right. But you have hijacked HaroTaro thread.
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Offline jeffchoi

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Re: Polyprotic buffers
« Reply #9 on: January 15, 2011, 08:35:41 PM »
how do i do the first step of neutralizing A3- to HA2-?

Offline Borek

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Re: Polyprotic buffers
« Reply #10 on: January 16, 2011, 05:38:22 AM »
A3- + H+ -> HA2-
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Offline eqsellah

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Re: Polyprotic buffers
« Reply #11 on: January 31, 2011, 12:22:24 PM »
Hi I am having trouble with this exact problem. In my original solution to the problem I used the pKa=6.9 for the buffer at pH=6.5, is this incorrect? Wouldn't it buffer more effectively? The reason I ask is because in the original post HaroTaro says they are using pH=2.1.

Also, I'm wondering if this can be treated as a monoprotic acid when using pKa=6.9 because the difference between pKas is quite large, the effect of the lowest pKa would be negligible?

Finally, my solution to the problem: How many moles of HCl should be added to prepare the buffer?

10^(6.5-6.9) = A- / HA
0.3981 = A- / HA

%HA = 71.53% * 0.2M = 0.14

0.14 moles would have to be added. Is this correct or am I missing something?

Offline Borek

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Re: Polyprotic buffers
« Reply #12 on: January 31, 2011, 03:15:37 PM »
Using pKa2 is OK, but your answer is wrong. You forgot to protonate AsO43- to HAsO42-.
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Offline eqsellah

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Re: Polyprotic buffers
« Reply #13 on: January 31, 2011, 11:15:24 PM »
Ok, then to protonate AsO43- to HAsO42- I would need an extra 0.2mol of HCl. To complete the buffer I would then add 0.14mol for a total of 0.34mol. Is this correct? Thanks for the *delete me*

Offline Borek

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Re: Polyprotic buffers
« Reply #14 on: February 01, 2011, 04:20:15 AM »
Much better now.
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