Salts in solution

  • Thread starter Big-Daddy
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  • #1
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Let's say I put SO2 or SO32- or HSO3- in solution. I get the following equilibria:

SO2 (aq) + H2O (l) ⇔ HSO3- (aq) + H3O+ (aq) ... K1
HSO3- (aq) + H2O (l) ⇔ SO32- (aq) + H3O+ (aq) ... K2

And this equilibrium always going on:

2H2O (l) ⇔ H3O+ (aq) + OH- (aq) ... Kw

Other than that, to solve the system we would write two more equations:
Mass Balance (S): Initial Concentration of S = [SO2] + [SO32-] + [HSO3-]
Charge Balance: [H3O+] = [OH-] + [HSO3-] + 2[SO32-]

So it looks like the only real initial variable is the Initial Concentration of S that we put in our system. Then, does this mean that if we put [itex]c[/itex] M of SO2 to begin with in the solution, the equilibrium concentrations of every species would be the same as if we had put [itex]c[/itex] M of Na2SO3 or [itex]c[/itex] M of NaHSO3? And also the same as if we put [itex](1/3)c[/itex] M of Al2(SO3)3? (as the initial concentration of S placed in the solution is the same one way or the other)
 

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  • #2
DrDu
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What about mass balance of H and O?
 
  • #3
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I don't know. How do you write mass balances for H and O?

In any case I've got 5 variables and 5 equations. That's besides my main point (though I am interested in what you just said too): please read the last paragraph of the post again.
 
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  • #4
DrDu
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No, you have 6 variables! While in a diluted solution it is justified to replace the activities [ ] with the corresponding concentrations for the dissolved substances, this is not true for the solvent, namely, [H2O]=1, but c(H2O)=n(H2O)/V.
So you have to take into account the mass balance of H (or of O, which is redundant):
2c(H2O)+3[H3O+]+[OH-]+[HSO3-]=const.
So the equilibrium depends not only on the amount of sulfur, but also of hydrogen or oxygen in the system.
 
  • #5
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First of all, what is this "constant"? The only value I'm fixing is the number of moles of sulphite added to the water, denoted "Initial Concentration of S". Other than that, you'll have to tell me what exactly the value for this constant would be, as used in the H mass balance.

Not only that, but what's to say my situation is not dilute enough for us to consider [H2O] as constant? That's the track I want to find out about. I'm not so worried at this point about H2O concentration being variable; I'm aware it's a necessary consideration but not before I get the answer to my main question. (Again, please read the last paragraph of the OP)
 
  • #6
DrDu
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The constant clearly is the total number of moles of Hydrogen in the volume considered.
So [itex]\mathrm{const.}=c_0(H)[/itex]. Compare this to your S being [itex] c_0(S)[/itex].
You can put in S in the form of SO2, SO32- and HSO3-. It is clear that adding HSO3- is changing the total amount of hydrogen present while adding SO2 or SO32- is not. Therefore the equilibrium concentrations reached will be different when you start from SO32- or SO2.
Not only that, but what's to say my situation is not dilute enough for us to consider [H2O] as constant?
The point is that in general for each species X you have to introduce two variables, namely the actual concentration c(X) and its activity a(X) or [X] in short.
For dilute solutions there is a simple relation between the two: [itex] a(X)\approx c(X)[/itex] and for solvents [itex]a(X)\approx 1[/itex].
While equilibrium constants contain activities, mass balance equations contain the true concentrations.
In not so dilute solutions, the relation between activities and concentrations can be quite complex.
For ionic substances, one can use e.g. Debye-Huckel theory:
http://en.wikipedia.org/wiki/Debye–Hückel_theory
 
  • #7
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Ok so that raises a few issues that I would love to go over again once my main question is answered. Did you just come here to raise my game from concentration to activity? I also had a question to start with...

Let's take the assumption that the solution is dilute enough for your activity to be 1 for the solvent and equal to concentration for the solutes.

In that case, the only real initial variable is the Initial Concentration of S that we put in our system. Then, does this mean that if we put c M of SO2 to begin with in the solution, the equilibrium concentrations of every species would be the same as if we had put c M of Na2SO3 or c M of NaHSO3? And also the same as if we put (1/3)c M of Al2(SO3)3? (as the initial concentration of S placed in the solution is the same one way or the other) Because in each case the same mass balance, charge balance and equilibrium relations should apply.

Please help me with this case. After that I will be more than happy to look at higher levels of precision and new forms of modelling that may be necessary for non-dilute solutions.
 
  • #8
DrDu
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I am desperately seeking to answer your question.
In that case, the only real initial variable is the Initial Concentration of S that we put in our system.
I am trying to convince you to see that this is wrong.
The total amount of hydrogen and oxygen in the system will be different depending on which of the three substances you add.
Let's say you have 1 mole of water, then you have 2 moles of hydrogen and one mole of oxygen.
If you add 0.1 moles of SO2 you have 2 moles of hydrogen and 1.2 moles of oxygen.
If you add 0.1 moles of HSO3- you have 2.1 moles of hydrogen and 1.3 moles of oxygen. (You have to consider also the counter ion in your charge balance)
If you add 0.1 moles of SO32- you have 2 moles of hydrogen and 1.3 moles of oxygen.
How can these three situations be equivalent?
 
  • #9
DrDu
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I just realized that the main point which is wrong in your argument is your charge conservation law.
Once you add, say NaHSO3, on the left hand side also [Na+] should show up.
 
  • #10
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Thanks. So if SO2 was added to the system, the equations are as I wrote above, if we can assume that the solution is dilute enough that activity = concentration for solutes and activity = 1 for the solvent.
Whereas if Na2SO3 or NaHSO3 is added to the system instead:

Charge Balance: [Na+] + [H3O+] = [OH-] + [HSO3-] + 2[SO32-]

But [Na+] is not a variable (assuming we add a known concentration of the salt, since Na+ will not react). Therefore, assuming the concentration of water to be unchanging and activity to be 1, we still have 5 variables and 5 equations to suit them - so what is the difference between adding Na2SO3 to the solution, and adding NaHSO3? Just that, if you had add c M of these, you'd have [Na+] = 2c, Initial Concentration of S = c for Na2SO3 and [Na+]=c, Initial Concentration of S = c for the NaHSO3 case, but no difference in the equations otherwise?

Once/if I get an understanding of the system without water's concentration being variable (because, let's face it, most of my textbook problems don't ask me to consider activity or water), I'll start to ask about water's activity being variable and try to understand that situation as well.
 
  • #11
DrDu
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Forget about the activity of water here. I was misguided.
Just that, if you had add c M of these, you'd have [Na+] = 2c, Initial Concentration of S = c for Na2SO3 and [Na+]=c, Initial Concentration of S = c for the NaHSO3 case, but no difference in the equations otherwise?
To me, that's a big difference.
 
  • #12
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To me, that's a big difference.
Well, the magnitude will show itself in trial problems :p But is that right? The only change we make to the equations used is, if we're using a salt instead of (say) SO2, add a term for [Na+] to the charge balance, where [Na+] = starting concentration of Na+ put in solution, since Na+ is not involved in any equilibrium reactions for this problem. The difference between Na2SO3 and NaHSO3 is as I pointed out in the last post, and only that. This is ok?

If so, I can't think what the difference would be between adding c M of SO2 and c M of H2SO3? (Assuming you can say that the first dissociation of H2SO3 goes to completion, which is assumed anyway when we say that there is no conversion from HSO3- to H2SO3 considered among our equilibria.)
 
  • #13
DrDu
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If so, I can't think what the difference would be between adding c M of SO2 and c M of H2SO3?
So can't I. But the situation will be completely different for e.g. Na2SO3 which is strongly alkaline, so in comparison with SO2 the resulting pH of the solution will be completely different.
 
  • #14
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So can't I. But the situation will be completely different for e.g. Na2SO3 which is strongly alkaline, so in comparison with SO2 the resulting pH of the solution will be completely different.
Right, so just a few things I want to check:

So long as we write the equilibrium equations for the system right, it doesn't really matter what we put in to start with. Thus SO2 is the same as H2SO3 but subtle differences in the equations in terms of multiplying [Na+] by a coefficient, between two salts, makes quite a large difference to the final result. A "buffer solution" is simply a weak acid or weak base to which a strong acid or strong base has been added, meaning that to modify the equations for a system with an acid and its buffer we need to include the metal ions from the buffer, and that's the only real difference to the equations (the presence of these cations in the charge balance causes other significant differences such as raising the alkalinity or reducing the acidity).

Salts are considered fully dissolved; if known otherwise, we're going to need to leave [Na+] as a variable and say that c0[Na2SO3] = [Na2SO3] + 2[Na+] and Ksp = [Na+]2[SO32-] or Ksp = [Na+][[HSO3-]. Alternatively we could use the same Ksp equation and c0[Na2SO3] = [Na2SO3] + [SO32-] + [SO2] + [HSO3-] (since all the concentration of these forms comes from the sulphite from the salt). Either way we've got two new variables ([Na+] and [Na2SO3]) and two new equations.

Or is it just that we forget the Na+ mass balance, and replace the equation which (previously) was c0[Na2SO3] = 2[Na+], with the equation Ksp = [Na+]2[SO32-] (same number of variables and equations as before)? What's wrong with the method above?

And lastly, in acid-base systems, how come we are not including the conjugate bases' alkaline equilibria? e.g. SO32- + H2O ⇔ HSO3- + OH-. Perhaps the equation turns out to be redundant because we have included both the Kw and Ka[HSO3-] equations in our system already?
 
  • #15
DrDu
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Perhaps the equation turns out to be redundant because we have included both the Kw and Ka[HSO3-] equations in our system already?
Yes, exactly. Just write down the three K's and see how one can be obtained from the other two.
 
  • #16
DrDu
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Now let's have some fun and consider another reaction equilibrium to show what I was first thinking off, namely, that you have sometimes have to take the concentration of H2O (via the mass balance of H or O) into account:
[itex]\mathrm{H_2O_2+SO_3^{2-}\leftrightarrow H_2O+SO_4^{2-}}[/itex]
Assume that all substances other than water are very diluted.
Apparently for this reaction
[itex]
K=\mathrm{\frac{[SO_4^{2-}]}{[H_2O_2][SO_3^{2-}]}}[/itex]
[itex] \mathrm{S=[SO_3^{2-}]+[SO_4^{2-}]}[/itex]
Total charge is 2S, so this does not add anything new. How do you proceed?
 
  • #17
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Happily, but can you please check what I wrote above that paragraph and clarify a few things first? I know it's a lot of reading but I wouldn't feel comfortable going on to something else until it's done and I feel ok with it. Text reproduced:

So long as we write the equilibrium equations for the system right, it doesn't really matter what we put in to start with. Thus SO2 is the same as H2SO3 but subtle differences in the equations in terms of multiplying [Na+] by a coefficient, between two salts, makes quite a large difference to the final result. A "buffer solution" is simply a weak acid or weak base to which a strong acid or strong base has been added, meaning that to modify the equations for a system with an acid and its buffer we need to include the metal ions from the buffer, and that's the only real difference to the equations (the presence of these cations in the charge balance causes other significant differences such as raising the alkalinity or reducing the acidity).

Salts are considered fully dissolved; if known otherwise, we're going to need to leave [Na+] as a variable and say that c0[Na2SO3] = [Na2SO3] + 2[Na+] and Ksp = [Na+]2[SO32-] or Ksp = [Na+][[HSO3-]. Alternatively we could use the same Ksp equation and c0[Na2SO3] = [Na2SO3] + [SO32-] + [SO2] + [HSO3-] (since all the concentration of these forms comes from the sulphite from the salt). Either way we've got two new variables ([Na+] and [Na2SO3]) and two new equations.

Or is it just that we forget the Na+ mass balance, and replace the equation which (previously) was c0[Na2SO3] = 2[Na+], with the equation Ksp = [Na+]2[SO32-] (same number of variables and equations as before)? What's wrong with the method above?
 
  • #18
DrDu
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Thus SO2 is the same as H2SO3 but subtle differences in the equations in terms of multiplying [Na+] by a coefficient, between two salts, makes quite a large difference to the final result.
I fear I don't understand most of what you have written. E.g. to which coefficient you are referring.
 
  • #19
DrDu
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Salts are considered fully dissolved; if known otherwise, we're going to need to leave [Na+] as a variable and say that c0[Na2SO3] = [Na2SO3] + 2[Na+] and Ksp = [Na+]2[SO32-] or Ksp = [Na+][[HSO3-]. Alternatively we could use the same Ksp equation and c0[Na2SO3] = [Na2SO3] + [SO32-] + [SO2] + [HSO3-] (since all the concentration of these forms comes from the sulphite from the salt). Either way we've got two new variables ([Na+] and [Na2SO3]) and two new equations.

Or is it just that we forget the Na+ mass balance, and replace the equation which (previously) was c0[Na2SO3] = 2[Na+], with the equation Ksp = [Na+]2[SO32-] (same number of variables and equations as before)? What's wrong with the method above?
This sounds basically ok. You can form various alternative equations by combining balance equations.
If a salt isn't completely dissociated you get two new variables, namely the concentrations of the undissociated salt and of the sodium ions. But you also have to consider two new equations: The mass action law for the dissociation related to the constant K and the mass balance for sodium.
 
  • #20
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I fear I don't understand most of what you have written. E.g. to which coefficient you are referring.
Ok, don't worry about that bit, I think it's ok. How about this:

A "buffer solution" is simply a weak acid or weak base to which a strong acid or strong base has been added, meaning that to modify the equations for a system with an acid and its buffer we need to include the metal ions from the buffer, and that's the only real difference to the equations (the presence of these cations in the charge balance causes other significant differences such as raising the alkalinity or reducing the acidity).

This sounds basically ok. You can form various alternative equations by combining balance equations.
If a salt isn't completely dissociated you get two new variables, namely the concentrations of the undissociated salt and of the sodium ions. But you also have to consider two new equations: The mass action law for the dissociation related to the constant K and the mass balance for sodium.
Hmm, there's a problem in my understanding.

It is said that salts in the aqueous phase then dissociate completely, no? In other words, K for reaction XA (aq) ⇔ X+ (aq) + A- (aq) is infinite. Ksp refers to the equilibrium XA (s) ⇔ X+ (aq) + A- (aq) instead. So then, it doesn't really make sense to include the concentration of XA, does it, since XA is a solid? But then how do I write a mass balance for sodium if the Na2SO3 salt is put in the solution and some is meant to have remained as solid? I would like to be able to write c0[Na2SO3] = [Na2SO3] + 2[Na+] but I can't because [Na2SO3] refers to a solid species. So the mass balance would be wrong, and we can't really write a mass balance that I can see?
 
  • #21
DrDu
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It is said that salts in the aqueous phase then dissociate completely, no? In other words, K for reaction XA (aq) ⇔ X+ (aq) + A- (aq) is infinite. Ksp refers to the equilibrium XA (s) ⇔ X+ (aq) + A- (aq) instead. So then, it doesn't really make sense to include the concentration of XA, does it, since XA is a solid? But then how do I write a mass balance for sodium if the Na2SO3 salt is put in the solution and some is meant to have remained as solid? I would like to be able to write c0[Na2SO3] = [Na2SO3] + 2[Na+] but I can't because [Na2SO3] refers to a solid species. So the mass balance would be wrong, and we can't really write a mass balance that I can see?
There are some exceptions like HgCl2, which is slightly soluble in water but dissociates only to a small extent.
If you have a not dissolved residue, you could in principle include it into your amount of moles balance
using [itex] \mathrm{2n_0(Na_2SO_3)=2n_s(Na_2SO_3) +V\cdot [Na^+]}[/itex] where V is the volume of your solution.
 
  • #22
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Ok but how do I finish my equation set if Na2SO3 (or some other salt which once dissolved can be assumed to dissociate completely as most salts are) is put into the solution and some precipitates out (I know this probably won't happen with Na2SO3 until quite a high limit, I'm asking in principle)? Do we just drop the mass balance - and write the Ksp equation? So then in the case of the original problem, I would have 6 variables ([Na+] is new) and now 6 equations as well?
 
  • #23
DrDu
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No, set up also the amount of substance balance for S. You can combine it with the one for Na so that the n's of the salt no longer appear. That's the equation you need.
 
  • #24
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You maybe don't understand what I'm asking. How can the Ksp not be needed to solve the problem, if I say that not all of the Na2SO3 dissolves? (By definition if not all the Na2SO3 dissolves then we are at saturation, so Ksp of Na2SO3 is valid to use.)
 
  • #25
DrDu
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Of course you need Ksp in this situation. I took this for granted, as you said so. I was trying to answer your question on whether to drop mass balance or not.
 

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