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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: HaroTaro on February 04, 2009, 06:18:47 AM

Title: Polyprotic buffers
Post by: HaroTaro 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.
Title: Re: Polyprotic buffers
Post by: Borek 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-.
Title: Re: Polyprotic buffers
Post by: HaroTaro 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?
Title: Re: Polyprotic buffers
Post by: aldoxime_amine 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.)
Title: Re: Polyprotic buffers
Post by: Borek on February 04, 2009, 05:16:53 PM
Quote from: aldoxime_amine on February 04, 2009, 03:05:09 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.
Title: Re: Polyprotic buffers
Post by: aldoxime_amine 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.
Title: Re: Polyprotic buffers
Post by: Borek 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.
Title: Re: Polyprotic buffers
Post by: aldoxime_amine 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?
Title: Re: Polyprotic buffers
Post by: Borek 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.
Title: Re: Polyprotic buffers
Post by: jeffchoi on January 15, 2011, 08:35:41 PM
how do i do the first step of neutralizing A3- to HA2-?
Title: Re: Polyprotic buffers
Post by: Borek on January 16, 2011, 05:38:22 AM
A3- + H+ -> HA2-
Title: Re: Polyprotic buffers
Post by: eqsellah 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?
Title: Re: Polyprotic buffers
Post by: Borek on January 31, 2011, 03:15:37 PM
Using pKa2 is OK, but your answer is wrong. You forgot to protonate AsO43- to HAsO42-.
Title: Re: Polyprotic buffers
Post by: eqsellah 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*
Title: Re: Polyprotic buffers
Post by: Borek on February 01, 2011, 04:20:15 AM
Much better now.