Triprotic Acid Titration Exact Calculation

[epenguin]

I'll go with the approximate method (so the only two phosphate species we need to worry about are [HPO42-] and [H2PO4-]. You've clarified things a great deal in your last post, most importantly with the note that the ratio of concentrations is the same as the ratio of moles (which should have been obvious but for some reason wasn't!).

We started with P_tot mol H₃PO₄ acid. So P_tot=n[H₂PO₄^-]+n[HPO₄^2-] and from the ratio equation we get P_tot=n[H₂PO₄^-]+(K_a2*n[H₂PO₄^-]/[H⁺]).

Now we solve for n[H₂PO₄^-]=P_tot*[H⁺]/([H⁺]+K_a2). My worry previously was that we could not work with [H⁺] like this because this is a concentration and everything else is moles, but now I'll proceed with the realization that because the ratios of concentration and moles are the same it makes no difference. n[H₂PO₄^-]=0.2*0.250*(10^(-7.80))/((10^(-7.80)+(10^(-7.20))=0.010038 moles roughly. n[HPO₄^2-]=0.039962 mol by subtraction.

How do we proceed to calculate n[K⁺]? Although we could easily use the charge balance in terms of number of moles, we would still be lacking n[H⁺]. We have n[H⁺]+n[K⁺]=0.010038+2*0.039962. But this must be the wrong route as without V we cannot work out n[H⁺] from [H⁺].

Not got time to check through that right now but is looking better. Try make it as simple as possible.
About the K⁺ and H⁺ I've written all needed in several posts above.

 

[Big-Daddy]

Not got time to check through that right now but is looking better. Try make it as simple as possible.
About the K⁺ and H⁺ I've written all needed in several posts above.

I'm afraid I've read through all your posts again and it's still not clear to me how to go from the number of moles of the two phosphate forms to the number of moles of K⁺.

 

[epenguin]

I
How do we proceed to calculate n[K⁺]? Although we could easily use the charge balance in terms of number of moles, we would still be lacking n[H⁺]. We have n[H⁺]+n[K⁺]=0.010038+2*0.039962. But this must be the wrong route as without V we cannot work out n[H⁺] from [H⁺].

That is absolutely the right idea and sort of formula (not checked the numbers).

How much is [H⁺] at pH 7.8?

I'm afraid I've read through all your posts again and it's still not clear to me how to go from the number of moles of the two phosphate forms to the number of moles of K⁺.

you just did it above.

You haven't yet given an idea of [H₃PO₄], requested in the problem and [PO₄^3-] desirable desirable to justify your simplifying assumption

 

[Big-Daddy]

That is absolutely the right idea and sort of formula (not checked the numbers).

How much is [H⁺] at pH 7.8?

you just did it above.

You haven't yet given an idea of [H₃PO₄], requested in the problem and [PO₄^3-] desirable desirable to justify your simplifying assumption

OK, once I've got the simplified method straight I will go back and calculate those two other concentrations additionally. Let me finish the problem this way first.

[H⁺] itself is 10^(-7.8)=1.58*10^-8 moldm^-3 but we currently have the moles of each phosphate form and the concentration of H⁺ - how do we get the moles of H⁺ (without the volume of the solution)?

 

[epenguin]

I have talked about [H⁺] in e.g. #4,18

I thought rather than repeating I would let you calculate it and draw you own conclusions.

The original phosphoric acid solution was 0.2M. It is not all that drastically diluted by the titration.

And you are worrying about the effect of something present at about 10^-8M ?

Edit: or more exactly at nearly 10^-6M, since that is [OH^-].

 

[Big-Daddy]

I have talked about [H⁺] in e.g. #4,18

I thought rather than repeating I would let you calculate it and draw you own conclusions.

The original phosphoric acid solution was 0.2M. It is not all that drastically diluted by the titration.

And you are worrying about the effect of something present at about 10^-8M ?

Edit: or more exactly at nearly 10^-6M, since that is [OH^-].

OK sorry I haven't been available over the last few days to talk this over, I was busy. Let me start from the beginning.

If we neglect [H₃PO₄] and [PO₄^3-], as the relevant pKa values are not within 3 units either way of the pH (7.8), we can write:

(Notation is that n[Species] is number of moles of that species. Working with this enables us not to worry about V when it comes to the conjugate base forms of the acid, as this V is not known since we don't know how much titrant was added.)

Mass balance for P: n[P_tot]=n[H₂PO₄^-] + n[HPO₄^2-]
Ratio of Conjugate Bases: n[HPO₄^2-]/n[H₂PO₄^-]=K_a2/[H⁺]
Charge Balance: n[K⁺]+n[H⁺]=n[H₂PO₄^-]+n[HPO₄^2-]+n[OH^-]

Now is the difficult bit and whatever I read of the post you just suggested (#18) I do not see how we can solve this. We do not know n[H⁺] or n[OH^-] so how can we work out n[K⁺]? Kw is useless because Kw=[H⁺]*[OH^-], not Kw=n[H⁺]*n[OH^-].

I'm not sure I understand your argument when it comes to dilution not having much effect. The moles of total acid are not going to change with dilution; meanwhile, given that we may have added nearly as much volume of KOH as we started with acid, I think it's ridiculous to say the volume hasn't changed at all.

Are you suggesting the approximation we make is n[H⁺]=0, n[OH^-]=0? Despite reading your last post again to myself I cannot see how this is justified except by calculating the values, and we cannot do that yet (if we could then there'd be no need to set them =0). If we could calculate n[H⁺] that would solve the problem on its own (as we just divide this by [H⁺] to find V, and the value of [H⁺] is known to be 10^(-7.8)).

Maybe if, instead of trying to tackle the various problems that are cropping up for me (as we clearly think in slightly different ways regarding this problem), you simply lay out your general technique for solving the problem (from the stage I reached of the equations I wrote above for mass balance, charge balance, etc.), with every step listed, I will understand where the solution to my own problems lies.

 

[epenguin]

Grrr as I had finished 3 para of answer I lost a connection and the post.

I will try to write what you ask in next days, but note it is unreasonable to imagine we can write chapters better than those of your book, so hopefully you will now see what you have now understood better is in your book.

The simplest justification for ignoring [OH^-] and [H⁺] here is that you are given the pH from which you can work out what they are and see that they are small and a negligible fraction of the charged species present.

 

[Big-Daddy]

Grrr as I had finished 3 para of answer I lost a connection and the post.

I will try to write what you ask in next days, but note it is unreasonable to imagine we can write chapters better than those of your book, so hopefully you will now see what you have now understood better is in your book.

The simplest justification for ignoring [OH^-] and [H⁺] here is that you are given the pH from which you can work out what they are and see that they are small and a negligible fraction of the charged species present.

My book says the answer begins with the Henderson-Hasselbalch equation, in this case we're using [HA^2-]/[H₂A^-]=K_a2/[H⁺], but other than that it gives no detail. I can assume that this is our starting point only because, as I can easily see now, the other species are negligible. My problems begin after I have succeeded in correctly calculating the moles of [HA^2-] and [H₂A^-] in solution.

I will be waiting for your response (whenever you have the time to write out the method). :)

By the way does the electroneutrality principle hold both in terms of concentrations and moles? In other words could we equally write a charge balance using moles as we could using concentration?

 

[epenguin]

By the way does the electroneutrality principle hold both in terms of concentrations and moles? In other words could we equally write a charge balance using moles as we could using concentration?

Certainly - any macroscopic part of the solution is neutral and the solution is neutral as a whole.

 

[Big-Daddy]

Certainly - any macroscopic part of the solution is neutral and the solution is neutral as a whole.

Thank you. I await your response to the rest of the question!

By the way I can confirm that the results for the number of moles of the two main species are practically the same with and without considering the other two (H₃PO₄ and PO₄^3-).