Triprotic Acid Titration Exact Calculation

[Big-Daddy]

Homework Statement

Calculate the analytical concentrations C₂[KOH] and C₂[H₃PO₄] in a solution where H₃PO₄ (originally of concentration 0.2 moldm^-3 and volume 0.250 dm³) is titrated with KOH (originally of concentration 0.8 moldm^-3, unknown volume used) until the pH reaches 7.8.

Homework Equations

Mass balance: C₂[H₃PO₄]=[H₃PO₄]+[H₂PO₄^-]+[HPO₄^2-]+[PO₄^3-]
C₂[KOH]=[K⁺]
Charge balance: [H⁺]+[K⁺]=[H₂PO₄^-]+2*[HPO₄^2-]+3*[PO₄^3-]+[OH^-]
Equilibria:
K_w=[H⁺]*[OH^-]
K_a1=[H₂PO₄^-]*[H⁺]/[H₃PO₄]
K_a2=[HPO₄^2-]*[H⁺]/[H₂PO₄^-]
Ka₃=[PO₄^3-]*[H⁺]/[HPO₄^2-]

The Attempt at a Solution

The unknowns are: [OH^-], [PO₄^3-], [HPO₄^2-], [H₂PO₄^-], [H₃PO₄] and [K⁺]. These number 6; we have 6 equations. So a solution exists. [OH^-]=K_w/[H⁺], but what about the rest?

My problem is that C₂[H₃PO₄] and C₂[KOH] are unknown of themselves. They represent the analytical concentrations of H3PO4 and KOH in the final solution (of pH=7.90) after they have been added together. These are unfortunately the two values we need to work out in the end.

 

[Sunil Simha]

Assume V as the final volume of the solution. Then C₂[H₃PO₄] would be equal to n/V where n is the initial number of moles of H₃PO₄.

Now write the anion conc. in terms of the dissociation constants and [H₃PO₄] plug these in the mass equation to get [H₃PO₄] in terms of V.

Now in the charge balance equation, [K⁺] is equal to 0.8*(V-0.25)/V, the anion concentrations are also in terms of V and [H₊] is known. Thus V can be found and hence the analytical concentrations.

 

[Big-Daddy]

Assume V as the final volume of the solution. Then C₂[H₃PO₄] would be equal to n/V where n is the initial number of moles of H₃PO₄.

Now write the anion conc. in terms of the dissociation constants and [H₃PO₄] plug these in the mass equation to get [H₃PO₄] in terms of V.

Now in the charge balance equation, [K⁺] is equal to 0.8*(V-0.25)/V, the anion concentrations are also in terms of V and [H₊] is known. Thus V can be found and hence the analytical concentrations.

Can you demonstrate how to write the anion concentrations in terms of dissociation constants? It's this step I always have trouble with for this type of calculation (i.e. the tough step is combining the mass balance and charge balance equations; I can do it sometimes, but not usually for triprotic/diprotic acids).

 

[epenguin]

Homework Statement

Calculate the analytical concentrations C₂[KOH] and C₂[H₃PO₄] in a solution where H₃PO₄ (originally of concentration 0.2 moldm^-3 and volume 0.250 dm³) is titrated with KOH (originally of concentration 0.8 moldm^-3, unknown volume used) until the pH reaches 7.8.

Homework Equations

Mass balance: C₂[H₃PO₄]=[H₃PO₄]+[H₂PO₄^-]+[HPO₄^2-]+[PO₄^3-]
C₂[KOH]=[K⁺]
Charge balance: [H⁺]+[K⁺]=[H₂PO₄^-]+2*[HPO₄^2-]+3*[PO₄^3-]+[OH^-]
Equilibria:
K_w=[H⁺]*[OH^-]
K_a1=[H₂PO₄^-]*[H⁺]/[H₃PO₄]
K_a2=[HPO₄^2-]*[H⁺]/[H₂PO₄^-]
Ka₃=[PO₄^3-]*[H⁺]/[HPO₄^2-]

The Attempt at a Solution

The unknowns are: [OH^-], [PO₄^3-], [HPO₄^2-], [H₂PO₄^-], [H₃PO₄] and [K⁺]. These number 6; we have 6 equations. So a solution exists. [OH^-]=K_w/[H⁺], but what about the rest?

My problem is that C₂[H₃PO₄] and C₂[KOH] are unknown of themselves. They represent the analytical concentrations of H3PO4 and KOH in the final solution (of pH=7.90) after they have been added together. These are unfortunately the two values we need to work out in the end.

Amounts of [H⁺] and [OH^-] are utterly negligible compared to other things in these conditions so just leave them out of eq. 2. [K⁺] is the same thing as [KOH]. OK after you have done your main calculation with molarities you finally express in terms of a volume.

Getting the requested equation is just a matter of applying the equilibrium equations you have written. You have to write your [H₂PO₄^-] in terms [H⁺] and K_a1, you practically have the expression for that bit, then you have to do it successively for the others and include them - you will get an [H⁺]² and [H⁺]³ term.

 

[Big-Daddy]

Amounts of [H⁺] and [OH^-] are utterly negligible compared to other things in these conditions so just leave them out of eq. 2. [K⁺] is the same thing as [KOH]. OK after you have done your main calculation with molarities you finally express in terms of a volume.

Getting the requested equation is just a matter of applying the equilibrium equations you have written. You have to write your [H₂PO₄^-] in terms [H⁺] and K_a1, you practically have the expression for that bit, then you have to do it successively for the others and include them - you will get an [H⁺]² and [H⁺]³ term.

Thanks, but in trying to get my [H₂PO₄^-] in terms of [H⁺] and K_a1 how do I get make [H₃PO₄] cancel down?

 

[Sunil Simha]

Thanks, but in trying to get my [H₂PO₄^-] in terms of [H⁺] and K_a1 how do I get make [H₃PO₄] cancel down?

Check out post #2.

 

[epenguin]

Thanks, but in trying to get my [H₂PO₄^-] in terms of [H⁺] and K_a1 how do I get make [H₃PO₄] cancel down?

Don't.
Don't worry for the moment and if we are taking it step by step don't ask me what to do after the step you haven't shown.

 

[Big-Daddy]

Don't.
Don't worry for the moment and if we are taking it step by step don't ask me what to do after the step you haven't shown.

According to your post this is the first step? You wrote: "You have to write your [H2PO4-] in terms [H+] and Ka1, you practically have the expression for that bit". In any case the problem still applies if I am to find [H₃PO₄] first: how do I get rid of [H₂PO₄^-] from the expression?

Perhaps if you could demonstrate for one of them, I will carry it on successively for the others.

 

[epenguin]

According to your post this is the first step? You wrote: "You have to write your [H2PO4-] in terms [H+] and Ka1, you practically have the expression for that bit". In any case the problem still applies if I am to find [H₃PO₄] first: how do I get rid of [H₂PO₄^-] from the expression?

Perhaps if you could demonstrate for one of them, I will carry it on successively for the others.

OK what I'm trying to get you to do is write all the species' concentrations in terms of [H₃PO₄] and at the end you will apply your mass balance equation. So just use your K_a1 equation to write [H₂PO₄^-] =
Very simple.

 

[Big-Daddy]

OK what I'm trying to get you to do is write all the species' concentrations in terms of [H₃PO₄] and at the end you will apply your mass balance equation. So just use your K_a1 equation to write [H₂PO₄^-] =
Very simple.

OK, so by substituting this successively we reach: [PO₄^3-]=[H₃PO₄]*K_a1*K_a2*K_a3/([H+]³). Now what?

 

[epenguin]

OK, so by substituting this successively we reach: [PO₄^3-]=[H₃PO₄]*K_a1*K_a2*K_a3/([H+]³). Now what?

OK you've got the last one - have you got the first one that I asked? And the second one?
(for [HPO₄^2-] = ?)

Write them down too please.

 

[Big-Daddy]

OK you've got the last one - have you got the first one that I asked? And the second one?
(for [HPO₄^2-] = ?)

Write them down too please.

[H₂PO₄^-]=[H3PO4]*Ka1/[H+]

[HPO₄^2-]=[H3PO4]*Ka1*Ka2/[H+]²

[PO₄^3-]=[H3PO4]*Ka1*Ka2*Ka3/([H+]³).

I'm beginning to see where this is going. Next we're going to substitute in for each term in the mass balance?

 

[epenguin]

[H₂PO₄^-]=[H3PO4]*Ka1/[H+]

[HPO₄^2-]=[H3PO4]*Ka1*Ka2/[H+]²

[PO₄^3-]=[H3PO4]*Ka1*Ka2*Ka3/([H+]³).

I'm beginning to see where this is going. Next we're going to substitute in for each term in the mass balance?

Yes. But give us the result.

 

[Big-Daddy]

Yes. But give us the result.

C₂[H₃PO₄]*[H⁺]³=[H₃PO₄]*([H⁺]³+Ka1*[H⁺]²+Ka1*Ka2*[H⁺]+Ka1*Ka2*Ka3)

And now what? We have neither C₂[H₃PO₄] and [H₃PO₄].

 

[epenguin]

C₂[H₃PO₄]*[H⁺]³=[H₃PO₄]*([H⁺]³+Ka1*[H⁺]²+Ka1*Ka2*[H⁺]+Ka1*Ka2*Ka3)

And now what? We have neither C₂[H₃PO₄] and [H₃PO₄].

You do have C₂[H₃PO₄], it's in the second line of the thread.

[H₃PO₄] is what you are trying to find the formula for. You have everything else and could now write

[H₃PO₄] = ...

Maybe you don't recognise what an answer would be, and in general it would be

[H₃PO₄] = C₂[H₃PO₄] X (function of the K's and [H⁺])

and that is the important one. In this case C₂[H₃PO₄] can be given a number that they want you to work out.

 

[Big-Daddy]

You do have C₂[H₃PO₄], it's in the second line of the thread.

[H₃PO₄] is what you are trying to find the formula for. You have everything else and could now write

[H₃PO₄] = ...

Maybe you don't recognise what an answer would be, and in general it would be

[H₃PO₄] = C₂[H₃PO₄] X (function of the K's and [H⁺])

and that is the important one. In this case C₂[H₃PO₄] can be given a number that they want you to work out.

No, we have C₁[H3PO4] (the original analytical concentration). C₂[H3PO4] is the analytical concentration of H3PO4 once H3PO4 was combined with a certain unknown volume of NaOH to produce the solution of pH = 7.8. How do we find either C₂[H3PO4] or [H3PO4]?

 

[epenguin]

No, we have C₁[H3PO4] (the original analytical concentration). C₂[H3PO4] is the analytical concentration of H3PO4 once H3PO4 was combined with a certain unknown volume of NaOH to produce the solution of pH = 7.8. How do we find either C₂[H3PO4] or [H3PO4]?

Now you tell me! I'm sorry, this is the first time this C₁ has been mentioned and you have asked a question without ever defining what the terms you ask about meant (!) so I just went on assumption and how suchlike questions are treated always.

Never mind, it is always much the same thing and you can expect to meet similar questions again and again and they are always dealt with by the same routine. And you do have the information for the calc.

So I'll just write the various concentrations in rational form - molarities - what need is there for C₁, C₂ etc? There is a total of all forms of the phosphate which you are given the information to calculate, this will usually be known or knowable, and even when it isn't you still formulate the same equations in terms of it algebraically. So here I'll call it [P]_tot, actually it does equal the total molarity of phosphorus in the solution so the tot is just added mind label.

So I'll rewrite the last line of my last post as

[H₃PO₄] = [P]_tot X (function of the K's and [H⁺])

You can start by writing [P]_tot = function given by the mass balance equation with your equilibrium equations enabling you to write in terms on one concentration.
By which you can calculate [H₃PO₄] if you know (as you do) [P]_tot, [H⁺] and the K's.

Likewise you can get an equation for and calculate the concentrations of all the four forms of phosphate at any given pH.

 

[epenguin]

I have to answer this in more than one post because of connection and seeing screen probs.

Things you should realize:

The LHS of your mass balance equation should be, in my terms [P]_tot, which is what I assumed you meant.

Before you do the calculation of [H₃PO₄] I hope you'll realize that it's going to be quite tiny - of the order of less than 10^-5 of the main components at pH 7.8. [PO₄^3-] will be likewise negligible.

To solve the rest of your problem you are going to have to calculate [H₂PO₄^-] and [HPO₄^2-] preferably as fractions of [P]_tot, similar kind of calculation to that for [H₃PO₄].

The other part of your problem was to calculate [K⁺] . How are you going to do that? Well think, as well as mass balance what other kind of balance is there here? :wink In fact you already have an equation for [K^+].

I remind you we said [H⁺] and [OH^-] are negligible. I think what throws students is the fact that these seem all important, the questions are about them, and yet at the same time they are negligible though some of the equations involve them vitally, in others they are ignored. Think of them as reporters on the state of the system and on the balance between the big players there in mass.

In fact there are only two forms of phosphate that matter in this question H₂PO₄^- and HPO₄^2-

Therefore actually you can solve it using pretty much the theory you will have already done with monobasic acids. And in any case even the full theory for tri- or multi- protic acids is scarcely different from that you have already done for mono-, the only difference is that there are more terms. We often like to write the more general equations first, and then home down on any particular simplifications for particular cases. In the case that the pK's were closer together we'd have to keep the more general ones.

And I have to say you are taking days to grind out equations that people write in fewer minutes. Conservation equations and equilibrium equations + any simplifications, that's it. In chemistry and biochemistry you'll meet any number of equations like that for multiple equilibria, you'll see them just written down with minimal explanation, maybe definition of the eq. constants by equations, maybe a conservation equation or two, maybe not because considered obvious. You should see what all the terms and the pattern are saying.

I nay not have time next days to continue this, anyone is welcome to take over if nec.

Edit: other obscure terminology - I don't know what is meant by 'analytical' concentrations, unnecessary adjective?

 

[epenguin]

In my (rational and standard ) terminology your eq. of #14

[P]_tot[H⁺]³=[H₃PO₄]*([H⁺]³+Ka1*[H⁺]²+Ka1*Ka2*[H⁺]+Ka1*Ka2*Ka3)

appears correct.

You just need to express that as
[H₃PO₄] = ...

and answer the rest of the question.

 

[Big-Daddy]

I see now, thank you. :)

Since this method involves fairly long computation I was wondering if we could now work on an approximate shortcut. When I asked this elsewhere (not specifying the need for an exact solution) I was told that we need only consider the two species of the acid involved in a pKa value which is within 3 units either way of the pH (assuming no other pKa value is also within 3 units of the pH, which is true for phosphoric acid - pKa2 is the only pKa within 3 units of the pH). So we need only consider [HPO₄^2-] and [H₂PO₄^-] - makes sense so far. I was then told the first step is to calculate the [HPO₄^2-]/[H₂PO₄^-] ratio. This is fairly easy: [HPO₄^2-]/[H₂PO₄^-]=K_a2/[H⁺]. But do you have any idea how we might proceed from here?

Perhaps I should clarify what we're looking for. C₁ of any acid or base is the total initial concentration of that acid or base in the solution it was originally in (e.g. we're given 0.80 M for the base). C₂ of any acid or base is the "initial" (and by this I mean the "total" - this is the collected concentration of all the forms of the acid in the solution) concentration of that acid or base, in the new solution into which they have been placed. With that in mind, the values we now seek are C₂ for both the acid and base.

 

[epenguin]

I see now, thank you. :)

Since this method involves fairly long computation I was wondering if we could now work on an approximate shortcut. When I asked this elsewhere (not specifying the need for an exact solution) I was told that we need only consider the two species of the acid involved in a pKa value which is within 3 units either way of the pH (assuming no other pKa value is also within 3 units of the pH, which is true for phosphoric acid - pKa2 is the only pKa within 3 units of the pH). So we need only consider [HPO₄^2-] and [H₂PO₄^-] - makes sense so far. I was then told the first step is to calculate the [HPO₄^2-]/[H₂PO₄^-] ratio. This is fairly easy: [HPO₄^2-]/[H₂PO₄^-]=K_a2/[H⁺]. But do you have any idea how we might proceed from here?

Perhaps I should clarify what we're looking for. C₁ of any acid or base is the total initial concentration of that acid or base in the solution it was originally in (e.g. we're given 0.80 M for the base). C₂ of any acid or base is the "initial" (and by this I mean the "total" - this is the collected concentration of all the forms of the acid in the solution) concentration of that acid or base, in the new solution into which they have been placed. With that in mind, the values we now seek are C₂ for both the acid and base.

If you read back you'll see I said e.g. #18 only these two are significant quantitatively at this pH.
I would be surprised if they had set you to calculate the triprotic formula and had never done with you the simple monoprotic one to which it practically simplifies in this case. Check back course notes or textbook.

The general triprotic formula which you were asked is NOT any 'long' computation and you are almost there. You just need now to give the formula [H₃PO₄] =
*

Then it is easy to calculate its numerical value at the pH of the problem and you will see it is, as I said , tiny. Likewise you can calculate the value of the two species that are important now.

The principles of the simplified case about which you ask now are just the same. You have a ratio, call it A/B. From mass balance you have the sum (A + B). **

* (My guess at what is stopping you so long is the frequently observed fact, what students did in very elementary algebra five or ten years ago, they can't do or recognise any more when the same thing is called 'chemistry'.)

**I do draw the line I'm afraid at explaining how to work out A and B.

 

[Big-Daddy]

If you read back you'll see I said e.g. #18 only these two are significant quantitatively at this pH.
I would be surprised if they had set you to calculate the triprotic formula and had never done with you the simple monoprotic one to which it practically simplifies in this case. Check back course notes or textbook.

The general triprotic formula which you were asked is NOT any 'long' computation and you are almost there. You just need now to give the formula [H₃PO₄] =
*

Then it is easy to calculate its numerical value at the pH of the problem and you will see it is, as I said , tiny. Likewise you can calculate the value of the two species that are important now.

The principles of the simplified case about which you ask now are just the same. You have a ratio, call it A/B. From mass balance you have the sum (A + B). **

* (My guess at what is stopping you so long is the frequently observed fact, what students did in very elementary algebra five or ten years ago, they can't do or recognise any more when the same thing is called 'chemistry'.)

**I do draw the line I'm afraid at explaining how to work out A and B.

First part of your post:
My problem which still persists is that I can write this:

[H₃PO₄]=C_a2*F_A (F_A is the acid dissociation function you alluded to earlier)
So:

[H₃PO₄]=C_a2*[H+]³/([H+]³+[H+]²*K_a1+[H+]*K_a1*K_a2+K_a1*K_a2*K_a3)

But I don't have the value of Ca2 (the initial concentration of the acid, in the solution with both acid and base) as I don't know what volume I added to my original concentration. I have the total moles of acid (let's call it n[H₃PO₄], I don't see what's unreasonable about that) which stays constant during titration, but the concentration will have changed from C_a1=n[H₃PO₄]/V_a to C_a2=n[H₃PO₄]/(V_a+V_b). I simply do not understand how to compensate for this problem.

I come up against the same problem using the approximate method (as you say, the principles are the same). You write the mass balance as C_a2=[HPO₄^-]+[H₂PO₄^-] and thanks to the ratio I found in my earlier post it is easy to place the expression in terms of just 2 variables, but we know neither the total concentration of acid C_a2 nor either of the species concentrations. This is clearest when C_a2 is replaced with n[H₃PO₄]/(V_a+V_b), and you have two unknowns: V_b, and whatever acid species form you substituted for from the ratio. Trying to work out either of the two acid forms always seems to require C_a2, which we do not have since we don't know how much base was used to titrate the acid.

My basic algebra is fine.

Edit: The derivation of the species concentration in terms of [H+] is nice, I must thank you for showing me it. Unfortunately I can't see how to use it in the case of a titration (clearly the expression is the same, but the starting concentration of the acid is not anymore).

 

[epenguin]

First part of your post:
My problem which still persists is that I can write this:

[H₃PO₄]=C_a2*F_A (F_A is the acid dissociation function you alluded to earlier)
So:

[H₃PO₄]=C_a2*[H+]³/([H+]³+[H+]²*K_a1+[H+]*K_a1*K_a2+K_a1*K_a2*K_a3)

But I don't have the value of Ca2 (the initial concentration of the acid, in the solution with both acid and base) as I don't know what volume I added to my original concentration. I have the total moles of acid (let's call it n[H₃PO₄], I don't see what's unreasonable about that) which stays constant during titration, but the concentration will have changed from C_a1=n[H₃PO₄]/V_a to C_a2=n[H₃PO₄]/(V_a+V_b). I simply do not understand how to compensate for this problem.

I come up against the same problem using the approximate method (as you say, the principles are the same). You write the mass balance as C_a2=[HPO₄^-]+[H₂PO₄^-] and thanks to the ratio I found in my earlier post it is easy to place the expression in terms of just 2 variables, but we know neither the total concentration of acid C_a2 nor either of the species concentrations. This is clearest when C_a2 is replaced with n[H₃PO₄]/(V_a+V_b), and you have two unknowns: V_b, and whatever acid species form you substituted for from the ratio. Trying to work out either of the two acid forms always seems to require C_a2, which we do not have since we don't know how much base was used to titrate the acid.

My basic algebra is fine.

Edit: The derivation of the species concentration in terms of [H+] is nice, I must thank you for showing me it. Unfortunately I can't see how to use it in the case of a titration (clearly the expression is the same, but the starting concentration of the acid is not anymore).

Finally you stated the general formula for the concentrations of one of the species present in a phosphate solution created by adding some alkali to phosphoric acid. For the record in my terminology this is just

[H₃PO₄] = [P]_tot[H⁺]³/([H+]³+K_a1*[H+]²+K_a1*K_a2*[H+]+K_a1*K_a2*K_a3)

The other three species' concentrations can be given in just the same way. The denominator would be the same for all of them and the numerator different, one term that also appears in the denominator. Lots of equations of this general kind come up all the time, when you have as often multiple equilibria, the equilibrium equations are combined with (one or more) mass balance equations. It is not higher math but elementary algebra applied to a physico-chemical situation and laws. Almost accountancy.

If algebra is not the problem I wonder if a confusing terminology is.

Anyway you say the volume of KOH added is unknown. You can know. We talked above of [K⁺].Just imagine the KOH added was concentrated and effectively didn't change the volume. From the pH, the K's, work out the concentrations of the two important forms of phosphate. Hence work out what [K⁺] is. You know the concentration of KOH used, so you can know the volume added. Then you can correct for the dilution (a somewhat secondary point IMHO).

 

[Big-Daddy]

Finally you stated the general formula for the concentrations of one of the species present in a phosphate solution created by adding some alkali to phosphoric acid. For the record in my terminology this is just

[H₃PO₄] = [P]_tot[H⁺]³/([H+]³+K_a1*[H+]²+K_a1*K_a2*[H+]+K_a1*K_a2*K_a3)

The other three species' concentrations can be given in just the same way. The denominator would be the same for all of them and the numerator different, one term that also appears in the denominator. Lots of equations of this general kind come up all the time, when you have as often multiple equilibria, the equilibrium equations are combined with (one or more) mass balance equations. It is not higher math but elementary algebra applied to a physico-chemical situation and laws. Almost accountancy.

If algebra is not the problem I wonder if a confusing terminology is.

Anyway you say the volume of KOH added is unknown. You can know. We talked above of [K⁺].Just imagine the KOH added was concentrated and effectively didn't change the volume. From the pH, the K's, work out the concentrations of the two important forms of phosphate. Hence work out what [K⁺] is. You know the concentration of KOH used, so you can know the volume added. Then you can correct for the dilution (a somewhat secondary point IMHO).

There's no way we can assume the KOH added didn't change the volume, because we're given the concentration - 0.80 M - and this is certainly not enough to make the 0.2 M (originally) H3PO4 concentration seem negligible.

I'm wary of trying to post-correct for dilution when I imagine it will have an enormously serious effect. We could well be adding 0.050 dm³ (50 cm³), maybe slightly more, to a 0.250 dm³ solution.

Perhaps I am to use the charge balance to work out [K⁺], then this is equivalent to what I notated as C₂[KOH] in my original mass balance for K, and this enables us to work out the Vb used after which it all falls into place?

 

[epenguin]

There's no way we can assume the KOH added didn't change the volume, because we're given the concentration - 0.80 M - and this is certainly not enough to make the 0.2 M (originally) H3PO4 concentration seem negligible.

I'm wary of trying to post-correct for dilution when I imagine it will have an enormously serious effect. We could well be adding 0.050 dm³ (50 cm³), maybe slightly more, to a 0.250 dm³ solution.

Perhaps I am to use the charge balance to work out [K⁺], then this is equivalent to what I notated as C₂[KOH] in my original mass balance for K, and this enables us to work out the Vb used after which it all falls into place?

I think subtleties are coming up that don't affect the issue. Whatever you think dilution does you have been told the pH at the end of the procedure. That fixes the ratio of the two phosphate forms. (From that point you can reason in terms of total moles rather than concentration if you wish, may make it easier.) So you can get the moles of each form. That in turns fixes moles of K⁺ that have been added. I think that is all watertight. After that you can think about what this implies for final volumes and concentrations.

 

[Big-Daddy]

I think subtleties are coming up that don't affect the issue. Whatever you think dilution does you have been told the pH at the end of the procedure. That fixes the ratio of the two phosphate forms. (From that point you can reason in terms of total moles rather than concentration if you wish, may make it easier.) So you can get the moles of each form. That in turns fixes moles of K⁺ that have been added. I think that is all watertight. After that you can think about what this implies for final volumes and concentrations.

The part in bold is where you've lost me.

I have an expression like this for each species (using your notation):

[H₃PO₄] = [P]_tot[H⁺]³/([H+]³+K_a1*[H+]²+K_a1*K_a2*[H+]+K_a1*K_a2*K_a3)

What then is [P]_tot? How do we calculate that? This is our big question.

 

[epenguin]

The part in bold is where you've lost me.

I have an expression like this for each species (using your notation):

[H₃PO₄] = [P]_tot[H⁺]³/([H+]³+K_a1*[H+]²+K_a1*K_a2*[H+]+K_a1*K_a2*K_a3)

What then is [P]_tot? How do we calculate that? This is our big question.

OK maybe I didn't see your point because I wouldn't do it in a way that causes this problem.
Just do it by moles then as per my last post which appears simplest.

You could do it by concentrations defining a volume = (0.25 + V_KOH) or something (nothing's stopping you) but I don't think personally it is worth the trouble and that it distracts from essentials.

 

[Big-Daddy]

OK maybe I didn't see your point because I wouldn't do it in a way that causes this problem.
Just do it by moles then as per my last post which appears simplest.

You could do it by concentrations defining a volume = (0.25 + V_KOH) or something (nothing's stopping you) but I don't think personally it is worth the trouble and that it distracts from essentials.

Hmm but I don't have the moles of H⁺, only the concentration in solution.

I will make evident where this brings in a problem: P_tot[H3PO4]=n_eq[H2PO4^-]+n_eq[HPO4^2-] (neq is the number of moles at equilibrium of the species in the brackets) and K_a2/[H⁺]=[HPO4^2-]/[H2PO4^-] but this is in concentrations, so what do we do? And I'm not sure you can do this but if we simply convert each concentration to a number of moles we'll end up with K_a2/(V*[H⁺])=[HPO4^2-]/[H2PO4^-]. This means that our substitution of this ratio into the mass balance I wrote for P won't manage to exclude V.

 

[epenguin]

Hmm but I don't have the moles of H⁺, only the concentration in solution.

I will make evident where this brings in a problem: P_tot[H3PO4]=n_eq[H2PO4^-]+n_eq[HPO4^2-] (neq is the number of moles at equilibrium of the species in the brackets) and K_a2/[H⁺]=[HPO4^2-]/[H2PO4^-] but this is in concentrations, so what do we do? And I'm not sure you can do this but if we simply convert each concentration to a number of moles we'll end up with K_a2/(V*[H⁺])=[HPO4^2-]/[H2PO4^-]. This means that our substitution of this ratio into the mass balance I wrote for P won't manage to exclude V.

I think your teachers are less likely to notice your concern for accurate figures than your failure to come up with even any crude ones, and could take this for total lack of grasp! In a lot of these things exact concentrations don't matter nearly as much as proportions or ratios and also orders of magnitude.

It is really time you put some numbers on this fifteen-minute problem on which you are stuck for almost as many days, so you can do the following:

For the first question, you can calculate straightaway [H₃PO₄]/[ P]_tot at pH 7.8. If that is a very small number that's all you need to be able to say [H₃PO₄] is so small it can be ignored. Even without this concern for the volume change which if you ignored you'd be only about 10% out. Even if the concentration mattered, 10% more negligible is still negligible! But the calculation justifies instead of assumes that you can ignore it. Likewise calculate and state the same justification for [PO₄^3-].

Then calculate what is the ratio of the two remaining forms of phosphate at pH 7.8? (You only need [H⁺] for this - see your own equations*.) The ratio of the concentrations or molarities and the ratio of the amounts or moles are exactly the same, since they are in the same beaker. Hence how many moles of each phosphate species are there? Hence calculate how many moles of K⁺ are there? Hence you can calculate how many ml of KOH was added. Then you have all these volumes and accurate concentrations that you want.

* in your first post - not this one K_a2/(V*[H⁺])=[HPO4^2-]/[H2PO4^-] - where has that V come from?

 

[Big-Daddy]

I think your teachers are less likely to notice your concern for accurate figures than your failure to come up with even any crude ones, and could take this for total lack of grasp! In a lot of these things exact concentrations don't matter nearly as much as proportions or ratios and also orders of magnitude.

It is really time you put some numbers on this fifteen-minute problem on which you are stuck for almost as many days, so you can do the following:

For the first question, you can calculate straightaway [H₃PO₄]/[ P]_tot at pH 7.8. If that is a very small number that's all you need to be able to say [H₃PO₄] is so small it can be ignored. Even without this concern for the volume change which if you ignored you'd be only about 10% out. Even if the concentration mattered, 10% more negligible is still negligible! But the calculation justifies instead of assumes that you can ignore it. Likewise calculate and state the same justification for [PO₄^3-].

Then calculate what is the ratio of the two remaining forms of phosphate at pH 7.8? (You only need [H⁺] for this - see your own equations*.) The ratio of the concentrations or molarities and the ratio of the amounts or moles are exactly the same, since they are in the same beaker. Hence how many moles of each phosphate species are there? Hence calculate how many moles of K⁺ are there? Hence you can calculate how many ml of KOH was added. Then you have all these volumes and accurate concentrations that you want.

* in your first post - not this one K_a2/(V*[H⁺])=[HPO4^2-]/[H2PO4^-] - where has that V come from?

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⁺].

 

[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-).