Salts in solution


by Big-Daddy
Tags: salts, solution
DrDu
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#19
Aug22-13, 02:19 PM
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Quote Quote by Big-Daddy View Post
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.
Big-Daddy
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#20
Aug22-13, 03:59 PM
P: 335
Quote Quote by DrDu View Post
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).

Quote Quote by DrDu View Post
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?
DrDu
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#21
Aug22-13, 04:16 PM
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Quote Quote by Big-Daddy View Post
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.
Big-Daddy
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#22
Aug22-13, 05:54 PM
P: 335
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?
DrDu
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#23
Aug23-13, 01:15 AM
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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.
Big-Daddy
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#24
Aug23-13, 12:17 PM
P: 335
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.)
DrDu
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#25
Aug24-13, 01:23 PM
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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.
Big-Daddy
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#26
Aug24-13, 02:55 PM
P: 335
If you use Ksp, you shouldn't need any other equation. Let's say I've got a solution with Na2SO3 and NaHSO3 both put in, of which Na2SO3 (let's say) is in up to saturation (but NaHSO3 is still entirely dissolved at this point). Then my equations are:

Mass Balance for S: c0[Na2SO3] + c0[NaHSO3] = [SO2] + [HSO3-] + [SO32-]
Charge Balance: [H3O+] + [Na+] = [HSO3-] + 2[SO32-] + [OH-]
Equilibria: SO2 (aq) + H2O (l) ⇔ HSO3- (aq) + H3O+ (aq) ... K1
HSO3- (aq) + H2O (l) ⇔ SO32- (aq) + H3O+ (aq) ... K2
2H2O (l) ⇔ H3O+ (aq) + OH- (aq) ... Kw
Na2SO3 (s) ⇔ 2Na+ (aq) + SO32- (aq) ... Ksp

There are 6 equations there and 6 ([SO2], [HSO3-], [SO32-], [OH-], [H3O+], [Na+]) variables. This is solved, no need for a Na mass balance. So then, am I right to say that, if I ever have to move from saying "this salt dissolves completely in solution" to treating the case where "some of the salt precipitates in solution", the only change I need to make is throw out the old mass balance for Na (if the dissolution was complete we'd have written 2c0[Na2SO3] + c0[NaHSO3] = [Na+], but it's not so we have to throw this equation out) and include the Ksp equation instead?
DrDu
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#27
Aug25-13, 03:22 AM
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Quote Quote by Big-Daddy View Post
Mass Balance for S: c0[Na2SO3] + c0[NaHSO3] = [SO2] + [HSO3-] + [SO32-]
And how do you know c_0(Na2SO3) which must be the amount of dissolved Na2SO3 in this equation?
Big-Daddy
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#28
Aug25-13, 05:26 AM
P: 335
I see, good point. So what do you do then? I'll assume my other 5 equations are the correct ones for the 6 variables, to which one equation (the form of which you might suggest) would be added for a complete set.
DrDu
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#29
Aug25-13, 10:10 AM
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I'd take the conservation of Na to eliminate the dissolved c(Na2SO4) from the conservation of S.
Big-Daddy
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#30
Aug25-13, 11:40 AM
P: 335
What do you mean? Can you just write the one equation?

The problem with Na+ conservation is that Na+ is also contributed to by another salt.
DrDu
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#31
Aug25-13, 01:22 PM
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Come on, try to work out something yourself!
Quote Quote by Big-Daddy View Post
What do you mean? Can you just write the one equation?

The problem with Na+ conservation is that Na+ is also contributed to by another salt.
Of course, but ##c_0(NaHSO3)## is known as it is completely dissolved, at least, that's what you were assuming.
Big-Daddy
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#32
Aug25-13, 04:07 PM
P: 335
I'm sorry, I've tried and I don't know the answer.
DrDu
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#33
Aug26-13, 02:33 AM
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Set up the conservation equation for Na and use it to eliminate c0(Na2SO3)=n0(Na2SO3)/V, where n0(Na2SO3) is the total amount of dissolved Na2SO3) (i.e. c0(Na2SO3) is an additional variable in your system).
Big-Daddy
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#34
Aug26-13, 07:31 AM
P: 335
Ok, what if I then added K2SO3 up to saturation? Is it

Mass Balance for S: (1/2) * [K+] + (1/2) * ([Na+] - c0[NaHSO3]) + c0[NaHSO3] = [SO2] + [HSO3-] + [SO32-]


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