Finding pH of a buffer with polyprotic acid

[4LeafClover]

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

 

[Borek]

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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methods

 

[4LeafClover]

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?

 

[Borek]

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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[4LeafClover]

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?

 

[Borek]

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-).

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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[4LeafClover]

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?

 

[Borek]

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

 

[4LeafClover]

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.

 

[Borek]

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

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methods

 

[4LeafClover]

The salt is Na2HPO4 (sodium pyrophosphate).

 

[Borek]

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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methods

 

[4LeafClover]

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?

 

[Borek]

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.

So my reaction is 2 NaOH + H3PO4 --> Na2HPO4 + 2 H2O.

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.

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.

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, HPO₄^2-, and if there is enough base it can get neutralized as well.

So I guess that leaves me with using the Henderson Hasselbach equation. The acid/conjugate base pair is H3PO4 and Na2HPO4

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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[4LeafClover]

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

 

[Borek]

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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[4LeafClover]

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.

 

[Borek]

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?

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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methods

 

[4LeafClover]

Yes! I am definitely following you this time.

 

[Borek]

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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methods

 

[4LeafClover]

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.

 

[Borek]

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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[4LeafClover]

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.

 

[Borek]

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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methods

 

[4LeafClover]

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.

 

[Borek]

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-.

Close, but wrong. Please write reaction equation between H₂PO₄^- and NaOH. In what molar ratio do they react?

Extra Credit: If there were still unreacted NaOH left, it would neutralize as much HPO4^2- as possible - until excess base ran out.

Yes. And what will be produced?

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[4LeafClover]

H2PO4- + NaOH --> H2O + HPO4^2-

So they react in a 1:1 ratio. We produced 0.45 moles of H2PO4- in the reaction between NaOH and the first proton of H3PO4. Therefore, we have 0.45 moles of H2PO4- to work with and 0.8 - 0.45 = 0.35 moles of NaOH left to work with. Since they react in a 1:1 ratio, NaOH is the limiting reagent. Therefore we will use 0.35 moles of H2PO4- and 0.35 moles of NaOH to produce 0.35 moles of HPO4^2-. This leaves us with 0 moles NaOH and .10 moles of H2PO4- and the 0.35 moles of HPO4^2- which was produced.

If there were excess base left, it would react with HPO4^2- to produce PO4^3-

 

[Borek]

.10 moles of H2PO4- and the 0.35 moles of HPO4^2-

OK. Do you see your buffer?

This way - calculating amount of acid, amount of base, and checking where does the neutralization end, you can deal with any mixture - it may contain any combination of H₃PO₄, NaH₂PO₄, Na₂HPO₄, Na₃PO₄ and NaOH.

If there were excess base left, it would react with HPO4^2- to produce PO4^3-

Right. Hopefully you will now know how multiprotic acids react with bases.

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methods

 

[4LeafClover]

Wow, yes, I think I finally get it. I am going to have to go back and look at all we've talked about a couple of times though to make sure I've got it down for exams and such.

So since .10 moles of H2PO4- is my acid and the 0.35 moles of HPO4^2- is my base, my equation is going to be the Henderson Hasselbach using the ratio .35/.10 and the pKa of H2PO4-. pH = -log(6.2*10^-8) + log(.35/.10) which is 7.75 pH. Is this finally correct?

 

[Borek]

Yes. I got 7.74 but that is probably effect of using slightly different value for pKa2.

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[4LeafClover]

Wow, thanks! Hahaha, that was the most frustrating problem I've dealt with in a while, but it feels very good to have gotten through it. Thanks for all the help!