Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Big-Daddy on October 25, 2012, 08:48:48 PM
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I have a solution of H3PO4, 85% wt/wt purity, 1.69 gcm-3 density and with a FW of 98.00 (I suppose I'd better acknowledge now I don't know what either "wt/wt purity" or "FW" mean). The solution is 250cm3 and contains 3.48 cm3 of concentrated H3PO4. (Again, I don't know what this means either.) The pKa values are as follows:
pKa1 = 2.15
pKa2 = 7.20
pKa3 = 12.44
When this is titrated with 40 cm3 of 0.80 M NaOH (take this as strong), calculate the exact concentrations of all species in the solution.
I take this as Na3PO4, Na2HPO4, NaH2PO4, H3PO4, H2PO4-, HPO42-, PO43-, H+ (H3O+) and OH-. How would I even go about this? The difficult bit is predicting how much of each salt (Na3PO4, Na2HPO4, and NaH2PO4) will be produced, and how to calculate exactly how many moles of H3PO4 will be used up. Once I have the concentrations in new solution of the salts and the acid, it is easy to reach the exact [H3O+] (using proton condition approach).
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(I suppose I'd better acknowledge now I don't know what either "wt/wt purity" or "FW" mean).
Do you mean Weight/Weight Percent (% w/w)? If so then % w/w is the weight in grams of a solute per 100 g of solution.
FW is formula weight, which is the same as Molar Mass. Molar mass of H2) ~ 18 g/mol
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(I suppose I'd better acknowledge now I don't know what either "wt/wt purity" or "FW" mean).
Do you mean Weight/Weight Percent (% w/w)? If so then % w/w is the weight in grams of a solute per 100 g of solution.
FW is formula weight, which is the same as Molar Mass. Molar mass of H2) ~ 18 g/mol
Thanks for the definition of FW - I suppose it should have been obvious given the Mr of phosphoric acid.
I gave you the question as it was written down, so the problem dictates 85% wt/wt purity, but unless you have a reason to think this is different from % w/w they are clearly close enough to be the same.
So if the solution of concentrated H3PO4 is 3.48 cm3 and has 85% wt/wt purity, I can work out the starting concentration of phosphoric acid in the total solution (250 cm3) as follows:
Mass of Conc. H3PO4 = 1.69*3.48=5.8812g
Mass of H3PO4 in Conc Solution = 0.85*5.8812=4.99902g
Moles of H3PO4 in Solution = 4.99902/98=0.0510104 mol
Concentration of H3PO4 in Solution = 0.0510104/0.250=0.204 moldm-3
Is that correct?
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Concentration of H3PO4 in Solution = 0.0510104/0.250=0.204 moldm-3
Looks OK to me.
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Concentration of H3PO4 in Solution = 0.0510104/0.250=0.204 moldm-3
Looks OK to me.
Thanks - do you know how I might go about solving the rest of the problem? i.e. finding expressions for each of the other species in the solution?
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See http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
Equations 9.11-9.13 allow calculation of speciation in the solution of diprotic acid of known pH. For triprotic acid you would need to play a little bit with the derivation, but it is just a matter of following identical scheme. Sure, you need to calculate pH first - and I would say that's the more challenging part, as using Henderson-Hasselbalch equation will give only approximate result; H3PO4 is too strong for assumption H2PO4- is produced only in neutralization. Generally speaking you should solve equation 11.16 for this page: http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-solution - but as you can safely assume second and thoird dissociation steps don't play any role here, in fact equation 11.15 will be enough - and as the solution is acidic it can be simplified even more (one of teh terms is so low it can be ignored).
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See http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
Equations 9.11-9.13 allow calculation of speciation in the solution of diprotic acid of known pH. For triprotic acid you would need to play a little bit with the derivation, but it is just a matter of following identical scheme. Sure, you need to calculate pH first - and I would say that's the more challenging part, as using Henderson-Hasselbalch equation will give only approximate result; H3PO4 is too strong for assumption H2PO4- is produced only in neutralization. Generally speaking you should solve equation 11.16 for this page: http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-solution - but as you can safely assume second and thoird dissociation steps don't play any role here, in fact equation 11.15 will be enough - and as the solution is acidic it can be simplified even more (one of teh terms is so low it can be ignored).
I have seen this material before. In fact, when it comes to solving for pH, I would simply use the equation found in Robert de Levie's OCP on Acid-Base Equilibria. It is as exact as the one you linked me to, and almost identical. As for the concentration equation, it is available for an n-protic acid on Wikipedia.
My problem here is that when you add a base to an acid a salt is produced with the anion of the acid and the cation of the base, so shouldn't all such problems require you to consider the creation of this salt in your calculations? To do so, you must know what concentration of each salt (Na3PO4, Na2HPO4 and NaH2PO4) is produced in the solution. And before you know that, how can you get exact [H3O+ to use the concentration equations?
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Have you checked what Ca and Cb mean?
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As far as I know, they generally mean the starting concentration of the acid and base, but in this case Ca can mean the total starting concentration of all sources of the anion PO43- and Cb can be the total starting concentration of all of the sources the cation Na+. That doesn't really help, since we still don't know how to work out what these starting concentrations are individually.
I'm adding a certain amount of NaOH to pure H3PO4 - how do I figure out how much of each salt is produced? I can see now that Ca genuinely does refer to the starting concentration of the acid because any new salt formed capable of producing that anion can only add up to the same starting concentration. Meanwhile, I will take it without question that Cb retains the definition of the starting concentration of the base NaOH.
So essentially, when an acid (or multiple acids) is mixed with a base (or multiple bases), the salt(s) formed do not affect the pH of the solution; it can be calculated directly from the original starting concentrations of acid and base, as shown by the equation in the pH calculator used by ChemBuddy (as long as the new Ca and Cb in the solution formed by combining the acidic solution with the basic one is calculated from the previous Ca and Va, and Cb and Vb). Am I right?
That still doesn't solve the main part of this problem, though. I can find [H3O+] and [OH-] from the starting concentrations of acid and base, but how do I find Na3PO4, Na2HPO4, NaH2PO4, H3PO4, H2PO4-, HPO42-, and PO43-? I'd like to learn what's going on in terms of the equilibria (as I need to calculate these concentrations exactly rather than with approximations).
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As far as I know, they generally mean the starting concentration of the acid and base
No, not the starting concentrations but analytical concentrations of base and acid in the mixture.
I'm adding a certain amount of NaOH to pure H3PO4
And you can use this information to calculate Ca and Cb.
I can find [H3O+] and [OH-] from the starting concentrations of acid and base, but how do I find Na3PO4, Na2HPO4, NaH2PO4, H3PO4, H2PO4-, HPO42-, and PO43-? I'd like to learn what's going on in terms of the equilibria (as I need to calculate these concentrations exactly rather than with approximations).
Once you know pH you can use formulas for ion speciation that I pointed you to earlier.
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Once you know pH you can use formulas for ion speciation that I pointed you to earlier.
I'm not trying to work out the concentration of any ions - I'm trying to work out the concentration of salts formed. Individually. And this in turn, I imagine, will affect the concentration of the ions left in pure ionic form in solution after some are converted into salts of Na+.
I can easily work out the total concentration of H3PO4 removed, i.e. the total concentration of salt formed, but there are 3 different types of salt - how do I know what concentration of each will be formed?
As for the left over ions, I could believe from the calculator that in any mixture of acids and bases the salts formed have no relation to the concentrations of the ions, so once I have pH those ion concentration calculations will always hold true regardless of how many acids/bases are in the mixture.
But the problem is still - how much Na3PO4 is formed, how much Na2HPO4 is formed, how much NaH2PO4? If you could provide me with the formula for this one example in the necessary form, perhaps I could understand where you're getting it from?
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I'm not trying to work out the concentration of any ions - I'm trying to work out the concentration of salts formed.
Salts are made of ions, don't they?
At first you put your question as
calculate the exact concentrations of all species in the solution.
For me that means concentrations of individual ions, not salts.
Besides, concentrations of salts are impossible to calculate.
Imagine you have 1 L of 1 M solution of Na2HPO4. What is concentration of Na3PO4 in this solution? Of NaH2PO4 in this solution? And before you will say question doesn't make sense, as these salts are not present, think twice - 1M solution of Na2HPO4 can be for example prepared by dissolving 0.5 moles of Na3PO4 and 0.5 moles of NaH2PO4 in 1 L of solution, or 0.5 moles of Na2HPO4 and 0.25 moles of each other salt an so on. So not only these salts "are" present, but actually there are infinitely many answers to the question.
At the same time you can calculate equilibrium concentrations of all ions present - and this answer is unambiguous and unique.
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I'm not trying to work out the concentration of any ions - I'm trying to work out the concentration of salts formed.
Salts are made of ions, don't they?
At first you put your question as
calculate the exact concentrations of all species in the solution.
For me that means concentrations of individual ions, not salts.
Besides, concentrations of salts are impossible to calculate.
Imagine you have 1 L of 1 M solution of Na2HPO4. What is concentration of Na3PO4 in this solution? Of NaH2PO4 in this solution? And before you will say question doesn't make sense, as these salts are not present, think twice - 1M solution of Na2HPO4 can be for example prepared by dissolving 0.5 moles of Na3PO4 and 0.5 moles of NaH2PO4 in 1 L of solution, or 0.5 moles of Na2HPO4 and 0.25 moles of each other salt an so on. So not only these salts "are" present, but actually there are infinitely many answers to the question.
At the same time you can calculate equilibrium concentrations of all ions present - and this answer is unambiguous and unique.
OK, so how would I go about calculating these equilibrium concentrations? Do I only need the triprotic version of the equations on the website you linked to (i.e. the salt production has no effect on the equilibrium concentrations of each of these ions, only the starting concentration of the acid and base)?
As for the salts themselves, what if I then added more of the salt Na2HPO4 (after the acid-base reaction)? Do I not need to know the concentration of each of the salts present individually to create the expression needed for pH calculation?
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OK, so how would I go about calculating these equilibrium concentrations? Do I only need the triprotic version of the equations on the website you linked to (i.e. the salt production has no effect on the equilibrium concentrations of each of these ions, only the starting concentration of the acid and base)?
Yes, triprotic versions of these equations will be enough.
As for the salts themselves, what if I then added more of the salt Na2HPO4 (after the acid-base reaction)? Do I not need to know the concentration of each of the salts present individually to create the expression needed for pH calculation?
If any salt is added you can easily convert it to new concentrations of the acid and the base, after all adding a salt is not different from adding acid and base separately. As explained earlier, individual salts concentrations are irrelevant and ambiguous.
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OK, so how would I go about calculating these equilibrium concentrations? Do I only need the triprotic version of the equations on the website you linked to (i.e. the salt production has no effect on the equilibrium concentrations of each of these ions, only the starting concentration of the acid and base)?
Yes, triprotic versions of these equations will be enough.
As for the salts themselves, what if I then added more of the salt Na2HPO4 (after the acid-base reaction)? Do I not need to know the concentration of each of the salts present individually to create the expression needed for pH calculation?
If any salt is added you can easily convert it to new concentrations of the acid and the base, after all adding a salt is not different from adding acid and base separately. As explained earlier, individual salts concentrations are irrelevant and ambiguous.
I now see what you might be saying. Am I right to interpret this: regardless of how many reactions the H3PO4 undergoes, I do not need to consider the salts produced, because I can simply take the analytical concentration of H3PO4 from before (modified only for the volume of base or salt with which is titrated, not for anything else) and the analytical concentration of NaOH (in the new solution which is a mixture of both) - Ca and Cb respectively - and then, if new salt is added thereafter to the solution from the H3PO4-NaOH titration, I have to add a term for that salt alone but I do not have to consider the salts previously in solution and can continue to consider the Ca and Cb of H3PO4 and NaOH as before. In other words, for the first titration, of H3PO4 by NaOH, I can use Ca and Cb (as shown by the pH Calculator equation or any others), and then, let's say I add some Na2HPO4 to the resultant solution as the second titration - all I will need to calculate the pH is to find the pH of the first solution's titration and the volume of that solution (which can be found from Ca, Cb, Va, Vb and Ka/Kb values) and then titrate that with an equivalent expression for the salt Na2HPO4 - I still don't need to know how much salt was produced for using the pH calculator equation, as I can just refer back to the original Ca of H3PO4. How accurate is this?
Sorry it was so difficult to try and explain what I meant. I was trying my best.
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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').
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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 (Ca1) and then add Al2(H2PO4)3 (concentration Cs), then to work out the concentration of H2PO4-, HPO42- or PO43- I could use the triprotic speciation equations as before but with Ca from the equations replaced by Ca1+3*Cs, 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?
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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 (Ca1) and then add Al2(H2PO4)3 (concentration Cs), then to work out the concentration of H2PO4-, HPO42- or PO43- I could use the triprotic speciation equations as before but with Ca from the equations replaced by Ca1+3*Cs, 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.
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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 NH4[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?
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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).
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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 Ksp, would I? My general calculations are not designed to handle Ksp in addition to Ka, Kb and Kw, 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.
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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)
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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 Ksp, 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)52-, Al(OH)63- 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.
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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 HPO42- 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+), PO43-, HPO42-, 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 HPO42- and Ka3 to the loss of H+ from HPO42- to PO43-, 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 Cs. 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 Cb, use Cb=2*Cs (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 Ca with Ca=3*Cs (since the coefficient on the source of the anion PO43- 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 PO43-, HPO42-, H2PO4-, to H3PO4.
Does this method work to find the ionic concentrations?
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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.
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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 Cs*2 (where Cs is the analytical concentration of the salt, and the salt is Al2(HPO4)3) for the Cb in the equation for concentration of conjugate acids and Al(OH)3 concentration and Cs*3 for the Ca 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.
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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 Cs*2 (...) for the Cb
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 Kai and Ca (or Kbi and Cb) - and [H+], which we use as input data.
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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 Cs*2 (...) for the Cb
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 Kai and Ca (or Kbi and Cb) - and [H+], which we use as input data.
Thank you! You've been a great help on this topic. :)