July 04, 2022, 08:45:33 AM
Forum Rules: Read This Before Posting


Topic: How do I Prepare this Buffer?  (Read 23600 times)

0 Members and 1 Guest are viewing this topic.

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
How do I Prepare this Buffer?
« 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.

« Last Edit: January 21, 2008, 04:23:13 AM by mrlucky0 »

Offline Alpha-Omega

  • Full Member
  • ****
  • Posts: 693
  • Mole Snacks: +360/-231
  • Gender: Female
  • Physical Inorganic Chemist

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #2 on: January 21, 2008, 03:04:47 AM »
« Last Edit: January 16, 2012, 02:54:06 PM by Arkcon »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #3 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #4 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.


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #5 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #6 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?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #7 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #8 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


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #9 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #10 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? 
« Last Edit: January 21, 2008, 05:48:43 AM by mrlucky0 »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #11 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #12 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?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27121
  • Mole Snacks: +1761/-405
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: How do I Prepare this Buffer?
« Reply #13 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mrlucky0

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +0/-0
Re: How do I Prepare this Buffer?
« Reply #14 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-]

Sponsored Links