
#19
Aug2213, 02:19 PM

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P: 3,369

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
Aug2213, 03:59 PM

P: 335

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). 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 c_{0}[Na_{2}SO_{3}] = [Na_{2}SO_{3}] + 2[Na^{+}] but I can't because [Na_{2}SO_{3}] 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
Aug2213, 04:16 PM

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P: 3,369

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
Aug2213, 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?




#23
Aug2313, 01:15 AM

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P: 3,369

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
Aug2313, 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.)




#25
Aug2413, 01:23 PM

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P: 3,369

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.




#26
Aug2413, 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: c_{0}[Na2SO3] + c_{0}[NaHSO3] = [SO2] + [HSO_{3}^{}] + [SO_{3}^{2}] Charge Balance: [H_{3}O^{+}] + [Na^{+}] = [HSO_{3}^{}] + 2[SO_{3}^{2}] + [OH^{}] Equilibria: SO_{2} (aq) + H2O (l) ⇔ HSO_{3}^{} (aq) + H_{3}O^{+} (aq) ... K_{1} HSO_{3}^{} (aq) + H2O (l) ⇔ SO_{3}^{2} (aq) + H_{3}O^{+} (aq) ... K_{2} 2H_{2}O (l) ⇔ H_{3}O^{+} (aq) + OH^{} (aq) ... K_{w} Na2SO3 (s) ⇔ 2Na^{+} (aq) + SO_{3}^{2} (aq) ... K_{sp} There are 6 equations there and 6 ([SO2], [HSO_{3}^{}], [SO_{3}^{2}], [OH^{}], [H_{3}O^{+}], [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 2c_{0}[Na2SO3] + c_{0}[NaHSO3] = [Na^{+}], but it's not so we have to throw this equation out) and include the K_{sp} equation instead? 



#27
Aug2513, 03:22 AM

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#28
Aug2513, 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.




#29
Aug2513, 10:10 AM

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P: 3,369

I'd take the conservation of Na to eliminate the dissolved c(Na2SO4) from the conservation of S.




#30
Aug2513, 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. 



#31
Aug2513, 01:22 PM

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P: 3,369

Come on, try to work out something yourself!




#32
Aug2513, 04:07 PM

P: 335

I'm sorry, I've tried and I don't know the answer.




#33
Aug2613, 02:33 AM

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P: 3,369

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).




#34
Aug2613, 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+]  c_{0}[NaHSO3]) + c_{0}[NaHSO3] = [SO2] + [HSO_{3}^{}] + [SO_{3}^{2}] 


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