[fobbz]

I have a lab to plan which asks to make a buffer solution of pH = 7.45. We are to use 0.5M H2P04- and 1.0M NaOH. The solution must be prepared into a 250.0mL volumetric flask.

Now to start this problem off I have calculated the [HA]/[A-] ratio to be 0.5723. I calculated this using the equation [HA]/[A-] * Ka = 10^(-7.45)

The Ka is 6.2x10-8.

From here I really don't know what to do. I tried to use the ratio to find the moles before and after, but my answer was far far off. The actual answer comes to 90mL buffer and 28.7 NaOH.


When I take this and convert them over to mols and find the pH, I find it to be 7.21???

attatched is my attempt at understanding this.

Please help me I have a lab tomorrow at 930 and I'm really starting to freak out.

*{MOD edit: crop and re-host image}*

[Borek]

Name acid and its conjugate base (in other words - state clearly what is HA and what is A^-).

[fobbz]

NaH2PO4, so the acid was H2PO4^- and its conjugate base HPO4^2-

[Borek]

OK (note you can format these formulas properly almost the same way your formatted charges).

What is the source of HPO₄^2-?

Knowing the source - can you use the stoichiometry to calculate initial amount of H₂PO₄^- required and amount of NaOH?

[fobbz]

The source is 0.5M NaH₂PO4^-. How would I proceed to use stoic. to calculate H2PO4- ? Would I look at the Ka of the acid and use an ICE table type for this situation?

[Borek]

Not necessarily. Just think what is the reaction that happens when you add NaOH to the H₂PO₄^- solution. There are two steps, but ignore the second and assume the first one went to completion. (It is not exactly true, but true enough to solve the problem).

[fobbz]

Apologies for not replying for so long, I had 4 midterms to study for.

I however am trying to solve this problem again and I made some calculations in order to try to do so.

The buffer requires a pH change of 0.2 max when 0.02 Moles H₃O⁺ are added. However as you can see the pH drops far more substantially.

I don't understand this problem at all, could someone please help with the method to find the voluems, I can't get my head around it.

[Borek]

Now you have added another condition, so the problem becomes slightly different.

But it is still based on the same concept.

Assume neutralization reaction goes to completion.

Your initial buffer must have pH of 7.45. That gives you information about ratio of concentrations of H₂PO₄^- and HPO₄^2-.

When you add strong acid HPO₄^2- is protonated back to H₂PO₄^-. Again, you can assume reaction goes to completion. That gives you anther equation involving concentrations of [HPO₄^2-] and [H₂PO₄^-]. At this moment you should have two equations in two unknowns - just solve and you will know what teh concentrations should be.

Then it is just a matter of finding a way to dolute the original reagents.

[Arkcon]

fobbz:  could you also try to do a little photo cleanup when you post your work -- crop out portions of your desk, make sure the image is as small as possible, use less colors -- we need black and white, not shades of brightness of your paper cause by uneven lighting.  Like I did here:

[fobbz]

fobbz:  could you also try to do a little photo cleanup when you post your work -- crop out portions of your desk, make sure the image is as small as possible, use less colors -- we need black and white, not shades of brightness of your paper cause by uneven lighting.  Like I did here:

I apologize I uploaded the wrong file I had edited, and had not previewed my post.

@Borek,

Sorry for being unclear before. So following what you said, assuming that the neutralization goes to completion, and after thinking a while I came up with two equations which i believe to be right...


Obviously incorporating the Henderson-Hasselbach equation. Is this correct? Moving forward I suppose I would try to come up with a solution whose initial volumes etc after making the buffer would result with the concentrations found above.

[Borek]

Sorry for being unclear before. So following what you said, assuming that the neutralization goes to completion, and after thinking a while I came up with two equations which i believe to be right...

In general looks OK to me, with an obvious math error (7.45-0.2=7.25 and not 7.35). I have not checked further, but that's how I would approach the problem.

After you have x and y it is just a matter of stoichiometry and dilution.

[fobbz]

Oops, that was stupid.
Okay awesome I solved it. Thank you so much for your help.