[Borek]

If I understand you correctly - you are getting closer, but I feel like you are still partially wrong.

Looks to me like you have already accepted fact that if you have an initial mixture of sodium phosphates its exact composition in terms of salts used to prepare it doesn't matter - the only thing that matters is the analytical concentration of H₃PO₄ and NaOH (Ca and Cb respectively). That's OK.

However, it looks like you are still missing the fact that after adding another batch of salts to this solution you don't need to worry about these salts - all you need is to calculate new Ca' and Cb' concentrations - that is, analytical concentration of phosphoric acid (which is sum of the old acid present and the new acid added now) and analytical concentration of NaOH (again, total of what was in the solution plus what was added).

In other words - after adding more salts method of calculating speciation doesn't change at all, the only thing that changes are the analytical concentrations Ca and Cb (which are now Ca' and Cb').

[Big-Daddy]

If I understand you correctly - you are getting closer, but I feel like you are still partially wrong.

Looks to me like you have already accepted fact that if you have an initial mixture of sodium phosphates its exact composition in terms of salts used to prepare it doesn't matter - the only thing that matters is the analytical concentration of H₃PO₄ and NaOH (Ca and Cb respectively). That's OK.

However, it looks like you are still missing the fact that after adding another batch of salts to this solution you don't need to worry about these salts - all you need is to calculate new Ca' and Cb' concentrations - that is, analytical concentration of phosphoric acid (which is sum of the old acid present and the new acid added now) and analytical concentration of NaOH (again, total of what was in the solution plus what was added).

In other words - after adding more salts method of calculating speciation doesn't change at all, the only thing that changes are the analytical concentrations Ca and Cb (which are now Ca' and Cb').

I see - so if I were doing a titration where I have a "mixture of H3PO4 and NaOH" (in concentrations Ca1 and Cb1) as the analyte and Ca(OH)₂ as the titrant (concentration Cb2), I would not need to consider the concentrations of any salts - I could simply use these three concentration values (along with appropriate equilibrium constants) to find the [H3O+], bypassing the fact that any salts are produced at all.

What about the concentration of species? If I add NaOH to H₃PO₄ (C_a1) and then add Al2(H2PO4)3 (concentration C_s), then to work out the concentration of H₂PO₄^-, HPO₄^2- or PO₄^3- I could use the triprotic speciation equations as before but with Ca from the equations replaced by C_a1+3*C_s, with [H₃O⁺] being whatever the [H₃O⁺] is after the salt is added, and with the Ka values continuing to refer to H₃PO₄'s Ka values. Is that right?

[Borek]

I see - so if I were doing a titration where I have a "mixture of H3PO4 and NaOH" (in concentrations Ca1 and Cb1) as the analyte and Ca(OH)₂ as the titrant (concentration Cb2), I would not need to consider the concentrations of any salts - I could simply use these three concentration values (along with appropriate equilibrium constants) to find the [H3O+], bypassing the fact that any salts are produced at all.

Yes and no - using Ca(OH)₂ would mean precipitation of calcium phosphates which would mean much more difficult calculations. But yes, as long as the bases added are strong and are not producing precipitates, you can combine them all together as Cb.

What about the concentration of species? If I add NaOH to H₃PO₄ (C_a1) and then add Al2(H2PO4)3 (concentration C_s), then to work out the concentration of H₂PO₄^-, HPO₄^2- or PO₄^3- I could use the triprotic speciation equations as before but with Ca from the equations replaced by C_a1+3*C_s, with [H₃O⁺] being whatever the [H₃O⁺] is after the salt is added, and with the Ka values continuing to refer to H₃PO₄'s Ka values. Is that right?

Yes, although Al(OH)₃ is a weak base and it will make things much more complicated. But generally speaking - assuming once again you added phosphate of a strong base - that's the correct reasoning.

[Big-Daddy]

Yes and no - using Ca(OH)₂ would mean precipitation of calcium phosphates which would mean much more difficult calculations. But yes, as long as the bases added are strong and are not producing precipitates, you can combine them all together as Cb.

Let us assume for now that no precipitation occurs (i.e. the ions remain in solution whatever their concentration).

Yes, although Al(OH)₃ is a weak base and it will make things much more complicated. But generally speaking - assuming once again you added phosphate of a strong base - that's the correct reasoning.

In what sense will having the cation come from a weak base make things more complicated? I learnt how to create pH expressions (similar to the one from the pH calculator) for salts including cations such as NH_4[sup]+[/sup], which is a weak base cation. But as far as the salts go, do I have anything to consider? Or do I simply carry forward the Ca of H3PO4 and Cb of NaOH and then treat Al2(HPO4)3 as the titrant?

Perhaps if you could explain what must be considered if the salt cation from the base is weak that would help, e.g. focusing on what happens when the Al2(HPO4)3 is added (I'm not too worried about precipitation in Ca(OH)2 for now). Hydrolysis is not a problem for my calculations. Is that the only issue here?

[Borek]

When you have a weak base present, calculation of pH becomes more difficult, as pH becomes a function of the Kb values as well.

Speciation calculations - once you know pH - are identical, although if you have a weak base, you also have to do speciation calculations for the base. Approach will be similar and it is possible to derive formulas in almost exactly the same way they are derived for the acid.

Sadly, you are trying to run before knowing how to walk. Why don't you at least browse the pH lectures and see how the general approach is used everywhere? At the moment you seem to be treating every case separately, while there is a common, universal approach to these problems and they are all identical in a way (even if the final equations can be completely different).

[Big-Daddy]

When you have a weak base present, calculation of pH becomes more difficult, as pH becomes a function of the Kb values as well.

Speciation calculations - once you know pH - are identical, although if you have a weak base, you also have to do speciation calculations for the base. Approach will be similar and it is possible to derive formulas in almost exactly the same way they are derived for the acid.

Sadly, you are trying to run before knowing how to walk. Why don't you at least browse the pH lectures and see how the general approach is used everywhere? At the moment you seem to be treating every case separately, while there is a common, universal approach to these problems and they are all identical in a way (even if the final equations can be completely different).

Run before I walk? I already have a general approach that can solve for when the salt contains a weak base cation (the example my textbook used is the NH₄⁺). I'm not trying to treat each case separately; I'm trying to find every loophole in my general approach and make sure I have it plugged.

But having a weak base cation, like Ca2+ or Al3+, doesn't mean I would need the K_sp, would I? My general calculations are not designed to handle K_sp in addition to K_a, K_b and K_w, so I'd have to solve the system from scratch, but that's a problem for another day.

One last thing - when you say "if you have a weak base, you also have to do speciation calculations for the base", what do you mean? You need to find analogous formulae to the acid speciation calculations except for bases, so that you can calculate each of the conjugate acid concentrations? Or are you saying that the bases somehow have an effect on the speciation calculations for acids (other than changing the [H3O+], which is obvious)? I hope it's the former you mean, because I wouldn't understand the second one if it was true.

[Big-Daddy]

I still have one question, which I would be grateful if you could help me with.

If I have a solution of a salt with a weak base's cation, e.g. Al2(HPO4)3, how do I calculate the concentrations of each species in there? (Given the 3 Kb values of Al(OH)3 and the 3 Ka values of H3PO4, as well as the analytical concentration of the salt and the Kw)

[Borek]

Run before I walk? I already have a general approach

Sorry, but you don't have it. You still don't see that there is one approach that fits all. Your latest posts is a perfect example of that, as you are asking how to calculate things that can be calculated using exactly the same approach I am talking about from the very beginning.

But having a weak base cation, like Ca2+ or Al3+, doesn't mean I would need the K_sp, would I?

Only if there is a risk of precipitation.

One last thing - when you say "if you have a weak base, you also have to do speciation calculations for the base", what do you mean? You need to find analogous formulae to the acid speciation calculations except for bases, so that you can calculate each of the conjugate acid concentrations? Or are you saying that the bases somehow have an effect on the speciation calculations for acids (other than changing the [H3O+], which is obvious)? I hope it's the former you mean, because I wouldn't understand the second one if it was true.

Speciation of acid doesn't depend on the base (other than through pH). But just like there are four forms of the phosphoric acid present in the solution, there are several forms of each base present in the solution - and you should calculate concentration of each form using Kb values.

In the case of Al³⁺ situation is even more complicated, as depending on pH it can either precipitate as Al(OH)₃ or be present as Al³⁺, AlOH²⁺, Al(OH)₂⁺, Al(OH)₄^-, Al(OH)₅^2-, Al(OH)₆^3- plus some complexes with more than one Al cation.

If I have a solution of a salt with a weak base's cation, e.g. Al2(HPO4)3, how do I calculate the concentrations of each species in there? (Given the 3 Kb values of Al(OH)3 and the 3 Ka values of H3PO4, as well as the analytical concentration of the salt and the Kw)

Calculate pH first, then calculate speciation using the same approach we are talking about all the time.

[Big-Daddy]

Speciation of acid doesn't depend on the base (other than through pH). But just like there are four forms of the phosphoric acid present in the solution, there are several forms of each base present in the solution - and you should calculate concentration of each form using Kb values.

Yes, I see - this I've done before (in my practice, it seems that the equations needed are analogous to the speciation calculations for acids, with Ca replaced by Cb, Ka by Kb and [H3O+] by [OH-]).

Calculate pH first, then calculate speciation using the same approach we are talking about all the time.

Let me analyse what I think I know and then ask my doubts:

The salt Al2(HPO4)3 is assumed to dissolve entirely into solution (in the absence of a Ksp value for it). This leaves 2 Al³⁺ ions and 3 HPO4^2- ions for every mole of the salt dissolved. The ions for which I need concentration at equilibrium are therefore Al³⁺, Al(OH)²⁺, Al(OH)₂⁺, Al(OH)₃ (let us assume Kb1 refers to the loss of OH- from Al(OH)3 to Al(OH)₂⁺, Kb2 from Al(OH)₂⁺ to Al(OH)²⁺, and Kb3 from Al(OH)²⁺ to Al³⁺), PO4^3-, HPO₄^2-, H₂PO₄^-, and H₃PO₄ (let us assume Ka1 refers to the loss of H+ from H3PO4 to H2PO4^-, Ka2 refers to the loss of H+ from H2PO4^- to HPO4^2- and Ka3 to the loss of H+ from HPO4^2- to PO4^3-, just like in a solution of the normal acid H₃PO₄).

I assume, when you say that the same approach fits all, you mean the speciation equation will still solve this. Here is my interpretation of that. Please point out the mistakes where they are there, or tell me if this is correct.

Let the concentration of Al2(HPO4)3 be C_s. Use the equation as you would for a normal base (with Al(OH)3 the form which has all hydroxides, Al(OH)₂⁺ the form which has lost 1, Al(OH)₂⁺ the form which has lost 1, and Al(OH)²⁺ the form which has lost 2, so forth - i.e. just like we would calculate Al(OH)3 for a base, we use the same one of the speciation equations to find the concentration of Al(OH)3 for this solution), but for C_b, use C_b=2*C_s (since the coefficient of Al in the formula of this salt is 2). Plug that in along with the original Kb values in the correct order from the base Al(OH)₃, and with the calculated [OH-], and you can find the concentration of each species from Al³⁺, Al(OH)²⁺, Al(OH)₂⁺ to Al(OH)₃.

Next, use the equation as you would for a normal acid (i.e. to calculate [H₃PO₄] in this solution we want the same form of the speciation equation as used to calculate H₃PO₄ from a solution of that acid), but replacing C_a with C_a=3*C_s (since the coefficient on the source of the anion PO_43- is 3). Plug that in along with the original Ka values in the correct order from the acid H3PO4, and with the calculated [H3O+], and you can find the concentration of each species from PO4^3-, HPO₄^2-, H₂PO₄^-, to H₃PO₄.

Does this method work to find the ionic concentrations?

[Borek]

What you wrote looks a little bit cryptic to me, but it can be just my English failing me. I have a general feeling you are on the right track. Just remember it is easier to calculate speciation using overall dissociation constants, not stepwise ones.

[Big-Daddy]

What you wrote looks a little bit cryptic to me, but it can be just my English failing me. I have a general feeling you are on the right track. Just remember it is easier to calculate speciation using overall dissociation constants, not stepwise ones.

What I am proposing is that I treat finding the concentration of Al(OH)3 in the salt solution identical to finding the Al(OH)3 in a solution of Al(OH)3, and same for all other conjugate acids formed in the solution, whereas I treat finding the concentration of all conjugate bases of the acid H3PO4 with the same forms of equation (i.e. in exactly the same way) as I would find each of their concentrations if the solution was of the acid H3PO4, inputting the value of C_s*2 (where C_s is the analytical concentration of the salt, and the salt is Al2(HPO4)3) for the C_b in the equation for concentration of conjugate acids and Al(OH)3 concentration and C_s*3 for the C_a in the equation for concentration of conjugate bases and H3PO4 in the salt solution. When I refer to these equations, I mean the ionic speciation equations (you provided a link to the diprotic form of them a while back, for example). So essentially, I use the same equations, and put in the same values for Ka (if finding concentrations of a conjugate base of H3PO4) or Kb (if finding concentrations of a conjugate acid of Al(OH)3), for the concentration of the ions in the salt solution as I would if they were in a solution of the pure acid or base, but since they are in a salt solution I use the salt analytical concentration (and multiply it by the coefficient of the ion in the salt, e.g. 2 for Al and 3 for HPO4- in Al2(HPO4)3).

Is this accurate? Please don't hesitate to ask for further clarification on a point of what I mean if that would help.

[Borek]

What I am proposing is that I treat finding the concentration of Al(OH)3 in the salt solution identical to finding the Al(OH)3 in a solution of Al(OH)3 (...) inputting the value of C_s*2 (...) for the C_b

Yes. If you will follow the derivation of speciation formulas you will see they are based only on the mass balance and definitions of dissociation constants, that means all that matters are values of K_ai and Ca (or K_bi and Cb) - and [H⁺], which we use as input data.

[Big-Daddy]

What I am proposing is that I treat finding the concentration of Al(OH)3 in the salt solution identical to finding the Al(OH)3 in a solution of Al(OH)3 (...) inputting the value of C_s*2 (...) for the C_b

Yes. If you will follow the derivation of speciation formulas you will see they are based only on the mass balance and definitions of dissociation constants, that means all that matters are values of K_ai and Ca (or K_bi and Cb) - and [H⁺], which we use as input data.

Thank you! You've been a great help on this topic.