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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: mrlucky0 on January 21, 2008, 02:34:56 AM

Title: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 02:34:56 AM
The Problem:

Given 0.1 M  solutions of Na3PO4 and H3PO4, describe the preparation of 1 L of a phosphate buffer at a pH of 7.5 . What are the molar concentrations of the ions in the final buffer solution, including Na+ and H+?

Relevant Equations and constants:

pk1 of H3PO4 = 2.15
pk2 of H2PO4- = 7.20

HH equation: pH = pka + log ( [A-] / [AH] )

My Attempt:


Let vA = volume of H3PO4 to add
Let vB = volume of Na3PO4 to add

Then:

vA + vB = 1 L
or vB = 1 L - vA

Subbing this into HH equation:

pH = pka + log ( [A-] / [AH] )
7.5 = 2.15 + log ( (1-vA) / vA ) // (the two 0.1 M's cancels out)

When I solve this I get a ridiculous solution. I am pretty sure my set-up is wrong and I am not understanding something. Isn't the Na3PO4 is source of OH- ions? I think I also need to use the pk2 of phosphoric acid. Do H+ become twice dissociated from H3PO4? Can someone show me the correct setup? Thanks.

Title: Re: How do I Prepare this Buffer?
Post by: Alpha-Omega on January 21, 2008, 02:55:03 AM
How to Prepare Phosphate Buffers of Specific pH:

http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/preparing-buffers.shtml

Additionally:

http://www.science.smith.edu/departments/Biochem/Biochem_353/buffer.html
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 03:04:47 AM
How to Prepare Phosphate Buffers of Specific pH:

http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/preparing-buffers.shtml

Additionally:

http://www.science.smith.edu/departments/Biochem/Biochem_353/buffer.html

Oh wow, thanks! I'm disappointed in myself for not finding that.
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 03:23:09 AM
Check carefully what acid and conjugated base you need in your Hendersaon-Hasselbalch equation. You don't have these - so think, what happens in the solution when you mix phosphoric acid and sodium phopsphate. There is some stoichiometry involved.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 04:03:28 AM
Check carefully what acid and conjugated base you need in your Hendersaon-Hasselbalch equation. You don't have these - so think, what happens in the solution when you mix phosphoric acid and sodium phopsphate. There is some stoichiometry involved.

I get Na+ ions and phosphate 2- and 1-  ions ?

Looking at the website I figured out what I did wrong: I didn't use the appropriate Pka value. Now I have:

pH = pk2 + log([A-]/[HA])
7.5 = 7.2 + log (1-x)/x
==> x= .33

Is 330 mL is the amount of trisodium phosphate I need then?

The textbook solution is 557.7 mL of trisodium phosphate with 444.3 mL of phosphoric acid. I just can't figure out how this solution is gotten.

Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 04:13:10 AM
describe the preparation of 0.1 L of a phosphate buffer

The textbook solution is 557.7 mL of trisodium phosphate with 444.3 mL of phosphoric acid.

Since when 557.7 mL + 444.3 mL = 0.1 L ?

But it has nothing to do with the way of finding answer.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 04:24:01 AM
describe the preparation of 0.1 L of a phosphate buffer

The textbook solution is 557.7 mL of trisodium phosphate with 444.3 mL of phosphoric acid.

Since when 557.7 mL + 444.3 mL = 0.1 L ?

But it has nothing to do with the way of finding answer.

Whoops. That was a typo. It should be 1 L. It's be corrected. Can you point me in the right direction?
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 04:47:47 AM
First of all - when you mix phosphoric acid with phosphate you in fact mix relatively strong acid (H3PO4) with relatively strong base (PO43-). So they will react:

H3PO4 + PO43- -> H2PO4- + HPO42-

To calculate buffer composition you have to use Henderson-Hasselbalch equation - but it have to contain correct conjugated acid/base pair:

pH = pKa + log([HPO42-]/[H2PO4-])

Now - assume you mix x mL of H3PO4 solution and y mL of PO43- solution. For obvious reasons you know that

x + y = 1000 mL

Try to find out formulas for HPO42- and H2PO4- concentrations using x and y volumes (hint: that's where the stoichiometry will be used). Put these formulas into Henderson-Hasselbalch equation - and you have set of two equations in x and y. Solve for x and y and you are done.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 05:00:21 AM
First of all - when you mix phosphoric acid with phosphate you in fact mix relatively strong acid (H3PO4) with relatively strong base (PO43-). So they will react:

H3PO4 + PO43- -> H2PO4- + HPO42-

To calculate buffer composition you have to use Henderson-Hasselbalch equation - but it have to contain correct conjugated acid/base pair:

pH = pKa + log([HPO42-]/[H2PO4-])

Now - assume you mix x mL of H3PO4 solution and y mL of PO43- solution. For obvious reasons you know that

x + y = 1000 mL

Try to find out formulas for HPO42- and H2PO4- concentrations using x and y volumes (hint: that's where the stoichiometry will be used). Put these formulas into Henderson-Hasselbalch equation - and you have set of two equations in x and y. Solve for x and y and you are done.

7.5 = 7.2 + log (1-x) / (x)
=> x= .33 L of acid
y = .67 L base

Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 05:12:42 AM
7.5 = 7.2 + log (1-x) / (x)
=> x= .33 L of acid
y = .67 L base

No. This way you are ignoring stoichiometry of the neutralization reaction. Try to follow exactly what I wrote.

If you mix 1 mole of phosphoric acid with 1.5 mole of phopsphate - what will you have in the solution?
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 05:19:54 AM
7.5 = 7.2 + log (1-x) / (x)
=> x= .33 L of acid
y = .67 L base

No. This way you are ignoring stoichiometry of the neutralization reaction. Try to follow exactly what I wrote.

If you mix 1 mole of phosphoric acid with 1.5 mole of phopsphate - what will you have in the solution?

Sorry I'm really not following this.

So I know  x+y = 1 L

How do use stoichiometry to figure out the relationship? 
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 05:56:59 AM
If you take x mL of 0.1M phosphoric acid, you have 0.1x moles of this acid, right? Now you add y mL of 0.1M phosphate - so you used 0.1y moles. What happens then? They react. Phosphoric acid reacts neutralizes base (phosphate) generating HPO4- and being itself neutralized to H2PO4-. When amounts of acid and base mixed are identical, you have an equimolar mixture of H2PO4- and HPO42-. Any excess of base added then generates more HPO42- removing more H2PO4-. So concentration of H2PO4- is

[H2PO4-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out HPO42- concentration.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 06:08:47 AM
Quote
[H2PO4-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out HPO42- concentration.

Where did this 2 come from?
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 06:11:56 AM
Stoichiometry - 2 protons that have to be neutralized. 0.1*x - 0.1*y will be concentration of H3PO4 left.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 21, 2008, 06:18:03 AM
If you take x mL of 0.1M phosphoric acid, you have 0.1x moles of this acid, right? Now you add y mL of 0.1M phosphate - so you used 0.1y moles. What happens then? They react. Phosphoric acid reacts neutralizes base (phosphate) generating HPO4- and being itself neutralized to H2PO4-. When amounts of acid and base mixed are identical, you have an equimolar mixture of H2PO4- and HPO42-. Any excess of base added then generates more HPO42- removing more H2PO4-. So concentration of H2PO4- is

[H2PO4-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out HPO42- concentration.

I figure that the HPO42- concentration should be 0.1 M - [H2PO4-]
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 21, 2008, 06:33:34 AM
Substitute into Henderson-Hasselbalch equation and solve.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 22, 2008, 02:09:03 AM
Quote
[H2PO4-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out H2PO4- concentration.

Where did this 2 come from?

Borek, I did follow your instructions and obtained the correct solution. But I'm a little unclear about the stoichiometry. Which two protons are you referring to that are being neutralized? From the reaction equation, isn't there only a transfer of 1 proton? Or, does the 2 come from the fact that H2PO4- still has 2 protons left on it?

Can you clarify this part a bit more? Thanks, you've already been extremely helpful.
Title: Re: How do I Prepare this Buffer?
Post by: Borek on January 22, 2008, 03:32:53 AM
When you neutralize 1 proton you have H2PO4- in the solution. You have to move further, hence the 2nd.
Title: Re: How do I Prepare this Buffer?
Post by: AWK on January 22, 2008, 03:59:11 AM
In such a cases I teach my students to exchange protons (and Na+) stepwise:
H3PO4 + Na3PO4 = NaH2PO4 + Na2HPO4
Then, depending which reagent should be in excess
NaH2PO4 + Na3PO4 = 2Na2HPO4
or
H3PO4 + Na2HPO4 = 2NaH2PO4
The stoichiometry should be done in moles or in concentrations when all reagents are in the same volume.
Title: Re: How do I Prepare this Buffer?
Post by: mrlucky0 on January 22, 2008, 04:14:54 AM
When you neutralize 1 proton you have H2PO4- in the solution. You have to move further, hence the 2nd.

Borek, let me see if I've got this straight. The HH equation calls for: log ([H2PO4-] / [HPO4-2])

But we're really starting from H3PO4, and going to HPO4-2. Two protons are neutralized in this way.

If this reasoning was correct, I'm just glad I finally understood it. Unfortunately this concept was not very apparent to me. 

Edit:

In such a cases I teach my students to exchange protons (and Na+) stepwise:
H3PO4 + Na3PO4 = NaH2PO4 + Na2HPO4
Then, depending which reagent should be in excess
NaH2PO4 + Na3PO4 = 2Na2HPO4
or
H3PO4 + Na2HPO4 = 2NaH2PO4
The stoichiometry should be done in moles or in concentrations when all reagents are in the same volume.


Oh, that makes sense to me now.