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Adding Base To Buffer System

  • #1

Homework Statement



You have 400 mL of a solution of 0.20 M Na3PO4 buffer at pH 6.5. You add 0.7 g
of NaOH to this solution. What will be the new pH?


Homework Equations



I suppose vital givens will be:

pH = p_Ka + log[(A-)/(HA)]

p_Ka values for H3PO4 = 2.15, 6.78, 12.4



The Attempt at a Solution



Concerns that I have are:

1) Did I use the correct p_Ka value? (This is if the Henderson-Hasselbalch equation is of any use.)
2) Did I find each species’ mols accordingly?
3) As a weak triprotic acid/system is featured, have I setup and represented the involved species for final pH?

0.20 M Na3PO4 * 0.4 L Na3PO4 = 0.08 mol Na3PO4

Base will react fully with acid.

The question that I have before typing out my workings is the following.

At pH 6.5, is this valid to propose: [H3PO4] + [H2PO4 (-)] + [HPO4 (2-)] + [PO4 (3-)] = 0.08 mol of phosphate? I thought that perhaps the first two species would be predominant, or not negligible, at this pH.

Any advice is appreciated.

Thank you.
 

Answers and Replies

  • #2
Borek
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1) Did I use the correct p_Ka value? (This is if the Henderson-Hasselbalch equation is of any use.)
2) Did I find each species’ mols accordingly?
3) As a weak triprotic acid/system is featured, have I setup and represented the involved species for final pH?
Hard to say not knowing what you did.

Base will react fully with acid.
Good assumption. In some particular cases it won't be true, but as long as pKa of the acid is between pH 3 and 11 (or even 2 and 12) you don't have to worry. Note 12.4 is a little bit too high to be sure.

At pH 6.5, is this valid to propose: [H3PO4] + [H2PO4 (-)] + [HPO4 (2-)] + [PO4 (3-)] = 0.08 mol of phosphate?
It is valid regardless of the pH (assuming you started with 0.08 moles of phosphates - which is the case here).
 
  • #3
I am then supposed to find each species’ mol value at 6.5 as organized below before taking into consideration the reaction with base? For purposes as you have stated above, the last p_Ka value will be rounded to 12.0.

10^4.5 = [H2PO4 (-)]/[H3PO4] or 31623 = [H2PO4 (-)]/[H3PO4]

10^-0.28 = [HPO4 (2-)]/[H2PO4 (-)] or 0.52481 = [HPO4 (2-)]/[H2PO4 (-)]

10^-5.5 = PO4 (3-)]/[HPO4 (2-)] or 3.1623*10^-6 = [PO4 (3-)]/[HPO4 (2-)]

Thank you.
 
  • #4
Borek
Mentor
28,305
2,688
You have to tell us what you are doing. First you posted a question, and asked if your approach was right, without showing what you did. Now you have posted some calculations, probably in hope we will again try to read your mind and guess what you are doing and why. Sorry, it will not work. I can guess what you are trying to do, but I prefer to work on explicitly posted information, to not waste my time in case I guessed wrong.
 
  • #5
I apologize for sounding incomplete.

I was not sure that the assumption [H3PO4] + [H2PO4 (-)] + [HPO4 (2-)] + [PO4 (3-)] = 0.08 mol was correct. I typed my concerns as foresight or anticipation of where I might get stuck, and they did not reflect an actual approach that I practiced as I did not want to waste time in solving and getting the wrong pH value under false assumptions.

Now that you have cleared up that the above equation is true, I am trying to solve for pH step-by-step gradually instead of a single post.

This is my plan. I don’t know each species’ mol value, so I am trying to solve for each species, but I am stuck on solving for all four if I cannot assume a form is negligible. I am using the ratio of conjugate base and acid for each p_Ka so that I may substitute for a common species.

For example, using what I posted in my second post I have derived the equation


[H2PO4-]/ 31623 + [H2PO4-] + 0.52481*[H2PO4-] + [PO4 (3-)] = 0.08 mol

from

[H3PO4] + [H2PO4 (-)] + [HPO4 (2-)] + [PO4 (3-)] = 0.08 mol.

However, there are two variables instead of one.

I don’t know how to carry on from here.

Thanks for any advice. I hope I am clearer. (By the way, my intention is not for you to solve this problem for me. I want to solve this on my own without any help, but I am honestly stuck.)
 
  • #6
Borek
Mentor
28,305
2,688
You are making it unnecessarily more complicated than it is. At pH 6.5 you are close to pKa2, close enough that you don't have to worry about H3PO4 and PO43- (it can be easily proven - you already did it partially at least, showing how many times concentrations of respectively H2PO4- and HPO42- are larger). Calculate concentrations of H2PO4- and HPO42- assuming they are the only forms present in the solution and neutralization with NaOH went to completion, then plug them into the Henderson-Hasselbalch equation (you have already listed it) and you are done. No need for a full blown analysis and calculation of everything.
 

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