[mrlucky0]

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.

[Alpha-Omega]

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

[mrlucky0]

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.

[Borek]

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.

[mrlucky0]

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.

[Borek]

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.

[mrlucky0]

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?

[Borek]

First of all - when you mix phosphoric acid with phosphate you in fact mix relatively strong acid (H₃PO₄) with relatively strong base (PO₄^3-). So they will react:

H₃PO₄ + PO₄^3- -> H₂PO₄^- + HPO₄^2-

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

pH = pKa + log([HPO₄^2-]/[H₂PO₄^-])

Now - assume you mix x mL of H₃PO₄ solution and y mL of PO₄^3- solution. For obvious reasons you know that

x + y = 1000 mL

Try to find out formulas for HPO₄^2- 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.

[mrlucky0]

First of all - when you mix phosphoric acid with phosphate you in fact mix relatively strong acid (H₃PO₄) with relatively strong base (PO₄^3-). So they will react:

H₃PO₄ + PO₄^3- -> H₂PO₄^- + HPO₄^2-

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

pH = pKa + log([HPO₄^2-]/[H₂PO₄^-])

Now - assume you mix x mL of H₃PO₄ solution and y mL of PO₄^3- solution. For obvious reasons you know that

x + y = 1000 mL

Try to find out formulas for HPO₄^2- 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

[Borek]

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?

[mrlucky0]

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? 

[Borek]

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 HPO₄^2-. Any excess of base added then generates more HPO₄^2- 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 HPO₄^2- concentration.

[mrlucky0]

[H₂PO₄^-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out HPO₄^2- concentration.

Where did this 2 come from?

[Borek]

Stoichiometry - 2 protons that have to be neutralized. 0.1*x - 0.1*y will be concentration of H₃PO₄ left.

[mrlucky0]

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 HPO₄^2-. Any excess of base added then generates more HPO₄^2- 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 HPO₄^2- concentration.

I figure that the HPO₄^2- concentration should be 0.1 M - [H₂PO₄^-]