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Homework Help: Finding pH of a buffer with polyprotic acid

  1. Apr 17, 2010 #1
    Question:
    What is the pH of a solution made by mixing 0.30 mol NaOH, 0.25 mol Na2HPO4, and 0.20 mol H3PO4 with water and diluting to 1.00L?

    My work:
    So I found that the Ka value of H3PO4 is 7.5*10^-3 and the Ka value of HPO4 (found in Na2HPO4) is 4.2*10^-13. Because Ka of HPO4 is so small, I know that it won't contribute much to the concentration of H+, but don't I still need to consider it when figuring out the pH? Or can I just drop it and do my calculations like I am working only with H3PO4?
    I also know that since we are working with 1L of solution, the number of moles of each molecule used is equivalent to that molecule's molarity.
    Other than that, I am stumped because of the two Ka values. Anyone know how to work this problem
     
  2. jcsd
  3. Apr 17, 2010 #2

    Borek

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    What equation describes pH of bufffer?

    Strongest acid reacts with the strongest base first. Find limiting reagent and calculate concentrations of acid and conjugate base.

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  4. Apr 17, 2010 #3
    So I use the Henderson Hasselbalch equation and use the pKa of the strongest acid (H3PO4). I am disregarding Na2HPO4 in this buffer and so the limiting reagent is the .2 moles of H3PO4, but since the equation asks for concentration, not quantity, my equation is pH = -log(7.5*10^-3) + log (.3/.2)? For some reason I feel as though my base/acid ratio should be different...am I correct? If so, what am I doing wrong and what do I need to change?
     
  5. Apr 17, 2010 #4

    Borek

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    You can't ingore Na2HPO4. When mixed with H3PO4 it will react giving NaH2PO4.

    Perhaps we should start differently. Final mixture can be prepared by mixing just phosphoric acid and sodium hydroxide. Can you calculate how much of each should be used? Can you calculate - remembering that stronger acids are neutralized first - how much of each acid you will have in the mixture after neutralization ends?

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  6. Apr 18, 2010 #5
    First off, I just want to say sorry for not being so quick to understand this one, but thanks for sticking with me and helping me work through it.

    Well, after seeing how you reacted H3PO4 with Na2HPO4 I think that I am going to need the henderson hasslelbach equation in the end to calculate the amount of conjugate acid (H3PO4) and conjugate base (H2PO4-). I chose these two because H3PO4 is the stronger base and Na2HPO4 will only produce a negligible amount of its conjugate.

    So one mol OH- reacts with 1 mol H+. NaOH is a strong base so I have .30 moles of OH-. 0.20 moles of H3PO4 gives me 0.04 moles H+ (7.5*10^-3 = x^2/0.20) and 0.25 moles of Na2HPO4 gives me a negligible amount of H+ (3.24*10^-7 moles) and so this means that I will have about 0.26 moles of OH- left over after neutralization.

    Going with the Henderson Hasselbach way of solving the equation, my above calculations also show that I should have 0.04 moles of the conjugate base H2PO4-.

    But argh!! I am not sure how to synthesize all this information to solve for pH. Use Henderson Hasselbach to calculate the pH using the pKa value for H3PO4 and the ratio 0.04/0.20? Or do I need to do something with the fact that I still have leftover strong base?
     
  7. Apr 18, 2010 #6

    Borek

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    No, you need to calculate concentrations of acid and conjugate base first.

    0.25 mol Na2HPO4 means 0.25 mole of H3PO4 mixed with 0.5 mole of NaOH. Do you see why? Do you see how to combine this information with 0.30 mol NaOH and 0.20 mol H3PO4?

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  8. Apr 18, 2010 #7
    Wow, no, I do not understand how you got that whatsoever. I am sure that somewhere I learned the information to help me understand what you just did or else my professor would not have given me this problem, but I have no idea what I it is. I am really feeling lost on this problem and even with all the help you have given me, I have not been able to answer it. Maybe I just need it to be explained to me first before I can understand. Can you walk me through it?
     
  9. Apr 18, 2010 #8

    Borek

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    It is close to midnight here, so we don't have much time.

    If you mix 2 moles of NaOH with 1 mole of H3PO4, what do you get?

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    methods
     
  10. Apr 18, 2010 #9
    Wow, well thank you so much for helping me through this even though it is such a late night. You really are great help. I really appreciate you helping me figure this out...not just giving me the answer without helping me understand like others do because then I never understand how to do it on the next assignment or exam when I really need it!!

    Ok, so for the reaction of 2 moles NaOH and 1 mole H3PO4. Na+ is a spectator ion so I am working with 2 moles OH- and 1 mole H3PO4. 2 OH- + H3PO4 --> 2 H2O + HPO4^2-. So this means that H3PO4 is my acid and HPO4^2- is my conjugate base. The OH becomes neutralized in the form of H2O.
     
  11. Apr 18, 2010 #10

    Borek

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    What salt have you produced by mixing 2 moles of NaOH with 1 mole of phosphoric acid. Give its formula.

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  12. Apr 19, 2010 #11
    The salt is Na2HPO4 (sodium pyrophosphate).
     
  13. Apr 19, 2010 #12

    Borek

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    Formula is OK, but it is not pyrophosphate, it is disodium hydrogen phosphate.

    <rant>Pyrophosphoric acid is something completely different, its fomula is H4P2O7. How you can mix these thing is beyond my comprehension.</rant>

    Do you see connection between salt that was produced, amounts of reagents used, and question I asked earlier about amounts of sodium hydroxide and phosphoric acid that have to be mixed to obtain 0.25 mole of Na2HPO4?

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  14. Apr 19, 2010 #13
    So my reaction is 2 NaOH + H3PO4 --> Na2HPO4 + 2 H2O. The limiting reagent is NaOH because I only have .30 moles NaOH, thus I can only react .15 moles H3PO4 with the 0.30 moles NaOH. This is going to give me 0.15 moles of Na2HPO4. This means that I am going to have 0.05 moles unreacted/non-neutralized H3PO4 and 0.40 moles of Na2HPO4 between what was created with the NaOH and H3PO4 reaction and what was already mixed in the solution in the beginning. So I guess that leaves me with using the Henderson Hasselbach equation. The acid/conjugate base pair is H3PO4 and Na2HPO4. So for my ratio is it 0.40/0.05?
     
  15. Apr 19, 2010 #14

    Borek

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    Please don't hurry ahead, as you are making mistakes faster than I can correct them. Answer questions one by one when I ask them.

    That's correct. But it is NOT the reaction that occurs in the buffer solution from the question. As I told you - slow down, you will see where we are leading.

    It doesn't work this way. You don't end with Na2HPO4. There are consecutive neutralization reactions. First, the strongest acid is neutralized - that is, H3PO4. If there is excess base left, strongest acid left will be neutralized - that is, H2PO4-. There is also a third acid, HPO42-, and if there is enough base it can get neutralized as well.

    This is not an acid/conjugate base pair.

    So, let's get back. You have 0.25 mole of Na2HPO4. Can you tell how many moles of NaOH and how many moles of H3PO4 had to be mixed to produce 0.25 mole of Na2HPO4? Don't think about additional base and acid listed in the question, for now concentrate on the salt only.

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    Last edited: Apr 19, 2010
  16. Apr 19, 2010 #15
    Since one mol of H3PO4 reacts with 2 moles of NaOH to produce 1 mole of Na2HPO4, I would need 0.25 moles of H3PO4 and 0.50 moles NaOH to produce 0.25 moles of Na2HPO4
     
  17. Apr 19, 2010 #16

    Borek

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    Right. That means final solution can be prepared either by mixing 0.30 mol NaOH, 0.25 mol Na2HPO4, and 0.20 mol H3PO4 (as stated in the question) OR by mixng 0.30 mol NaOH, 0.50 mol NaOH, 0.25 mol of H3PO4, and 0.20 mol H3PO4.

    Can you see why?

    Now, how much NaOH, how much H3PO4, how far does the neutralization go (see my previous post)?

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  18. Apr 19, 2010 #17
    I don't fully understand, but here is what I do understand: Since there are consecutive neutralization reactions, as you said, the strongest acid will be neutralized first and the weakest acid neutralized last. The neutralization ends when all base has been neutralized (or acid I guess but in this problem, the base is the limiting reagent).

    But since NaOH is the limiting reagent, I don't understand how you can make the same solution using either NaOH and having unreacted H3PO4 and Na2HPO4 leftover OR by having nearly 3x the amount of NaOH and this time, neutralizing most all of the acids. To me, the first solution would have a pH below 7 and the second one would have a pH above 7.

    So could you please explain why it is that you could make the same solution with either 0.30 mol NaOH, 0.25 mol Na2HPO4, and 0.20 mol H3PO4 OR by mixing 0.30 mol NaOH, 0.50 mol NaOH, 0.25 mol of H3PO4, and 0.20 mol H3PO4? I am completely lost, and I don't feel like I have gotten any further in solving this problem than when we first started talking about it a couple of days ago. You're giving me all these obscure hints that don't make any sense to me, and I am no closer to understanding what I am doing than before. Obviously, you are trying to lead me to the answer, but what I really want is also to be able to understand, and I am not understanding at all right now. Can you just explain this process to me in full? Because what we are doing right now is not helping me learn, and I feel like if I saw how to do it, I would be able to understand why things were done the way they were.
     
  19. Apr 19, 2010 #18

    Borek

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    Let's split it into two parts and let's change order in which substances are listed.

    Solution contains 0.30 mol NaOH, 0.20 mol H3PO4 (these two are temporarilly not important) and 0.25 mol Na2HPO4, right?

    You can prepare 0.25 mol Na2HPO4 mixing 0.50 mol of NaOH and 0.25 mol H3PO4, right?

    All you do now is you replace 0.25 mol Na2HPO4 in the solution by reagents that can be used to prepare this exact amount of salt. That means, instead of putting 0.25 mol Na2HPO4, put 0.50 mol of NaOH and 0.25 mol H3PO4 there. Do you see that it is the same?

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  20. Apr 19, 2010 #19
    Yes! I am definitely following you this time.
     
  21. Apr 19, 2010 #20

    Borek

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    OK, let's get back to the solution containing 0.30 mol NaOH, 0.50 mol NaOH, 0.25 mol of H3PO4, and 0.20 mol H3PO4.

    How much NaOH (in total), how much H3PO4 (in total), how far does the neutralization go?

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  22. Apr 19, 2010 #21
    Well, when you ask how much NaOH and H3PO4, I am assuming you are asking how much is used up in the reaction. All of the 0.8 moles of NaOH will be used up. 0.4 moles of the H3PO4 will be used up. 0.05 moles of H3PO4 left. All of what goes into the reaction gets neutralized except for the 0.4 moles of Na2HPO4 that is produced by the reaction.
     
  23. Apr 19, 2010 #22

    Borek

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    Not exactly.

    0.8 mole of NaOH. 0.45 mole of H3PO4.

    Now recall what I wrote here about neutralization of multiprotic acid.

    You are right that all NaOH will be consumed, but let's split neutralization into steps.

    How much NaOH will be used to react with FIRST PROTON of H3PO4?

    What will be present after this first proton was neutralized?

    What will happen next?

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  24. Apr 19, 2010 #23
    0.45 moles of NaOH will react with the first proton of H3PO4. After the first proton is neutralized, there will be H2PO4- present. The excess base will neutralize as much H2PO4- as possible until all NaOH is used up.
     
  25. Apr 19, 2010 #24

    Borek

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    OK - so, how much H2PO4- will be neutralized (and what will be produced - and how much)? And how much H2PO4- will be left?

    And an extra credit question - what would happen if after neutralization of H2PO4- there would be still unreacted NaOH left?

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  26. Apr 20, 2010 #25
    0.175 moles of H2PO4- will be neutralized. There will be 0.275 moles H2PO4- left. This will produce 0.175 moles of HPO4^2-. Extra Credit: If there were still unreacted NaOH left, it would neutralize as much HPO4^2- as possible - until excess base ran out.
     
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