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Topic: simultaneous equilibria  (Read 11846 times)

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Offline curiouscat

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Re: simultaneous equilibria
« Reply #15 on: October 29, 2012, 12:15:30 AM »

Mathematica spits out three solutions for this system -- each two-pages long. None of them seems right to me.


Try adding constraints. Problem with Mathematica often is that it assumes the most general domain for x,y,z. Say, all complex.

Try adding:

 x,y,z ε Real
x,y,z all positive

x,y,z < A0, B0

etc.

Offline Big-Daddy

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Re: simultaneous equilibria
« Reply #16 on: October 31, 2012, 06:38:41 PM »
As I expected, solving these equations is not easy. Maxima already works for about half an hour and have not yet found the solution.

Is Maxima a software good at solving these problems (e.g. simultaneous non-linear equations, solve analytically for a particular variable in terms of chosen set of others, provided of course that such a solution is possible at all)? I was looking for a software that could help solve this type of problem.

Also, is an analytical solution always possible when the equilibrium constants involved are limited to Ka, Kb, Kw, Ksp and/or Kf?

Offline Borek

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Re: simultaneous equilibria
« Reply #17 on: October 31, 2012, 07:01:48 PM »
Is Maxima a software good at solving these problems (e.g. simultaneous non-linear equations, solve analytically for a particular variable in terms of chosen set of others, provided of course that such a solution is possible at all)? I was looking for a software that could help solve this type of problem.

From what I know Maxima is not best, and it has some quirks that I don't like. But it is free and reasonably good in my experience.

Quote
Also, is an analytical solution always possible when the equilibrium constants involved are limited to Ka, Kb, Kw, Ksp and/or Kf?

Quite the opposite, you can be sure there is no analytical solution in most cases, apart from the most simple ones.

Strong acid in water yields a second degree equation in [H+]. Each additional dissociation constant increases the degree by 1, so solution for a monoprotic weak acid will be a polynomial of a 3rd degree, of diprotic acid 4th degree, of triprotic acid - 5th degree, and it is a known fact that there is no analytical solution for 5th degree polynomials.
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Offline Big-Daddy

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Re: simultaneous equilibria
« Reply #18 on: October 31, 2012, 11:18:55 PM »
Is Maxima a software good at solving these problems (e.g. simultaneous non-linear equations, solve analytically for a particular variable in terms of chosen set of others, provided of course that such a solution is possible at all)? I was looking for a software that could help solve this type of problem.

From what I know Maxima is not best, and it has some quirks that I don't like. But it is free and reasonably good in my experience.

Quote
Also, is an analytical solution always possible when the equilibrium constants involved are limited to Ka, Kb, Kw, Ksp and/or Kf?

Quite the opposite, you can be sure there is no analytical solution in most cases, apart from the most simple ones.

Strong acid in water yields a second degree equation in [H+]. Each additional dissociation constant increases the degree by 1, so solution for a monoprotic weak acid will be a polynomial of a 3rd degree, of diprotic acid 4th degree, of triprotic acid - 5th degree, and it is a known fact that there is no analytical solution for 5th degree polynomials.

Yes, but to find the polynomial in [H+]? That should always be possible, no?

After that, substitution of numerical values into the extra terms (Ca, Ka, Kw, etc.) yields an equation that can always be solved so long as there is a real solution.

I will do some investigating into mathematical software that may be able to reach the polynomial stage from the set of simultaneous equations using abstract algebra alone (i.e. without specifying the values of the constants or concentrations, leaving them in algebraic form).

Offline curiouscat

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Re: simultaneous equilibria
« Reply #19 on: November 01, 2012, 12:55:01 AM »


Is Maxima a software good at solving these problems (e.g. simultaneous non-linear equations, solve analytically for a particular variable in terms of chosen set of others, provided of course that such a solution is possible at all)? I was looking for a software that could help solve this type of problem.

Mathematica is the best I've seen. Unfortunately not free.

Maple is decently good.

Offline Borek

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Re: simultaneous equilibria
« Reply #20 on: November 01, 2012, 05:09:54 AM »
Strong acid in water yields a second degree equation in [H+]. Each additional dissociation constant increases the degree by 1, so solution for a monoprotic weak acid will be a polynomial of a 3rd degree, of diprotic acid 4th degree, of triprotic acid - 5th degree, and it is a known fact that there is no analytical solution for 5th degree polynomials.

Yes, but to find the polynomial in [H+]? That should always be possible, no?

In my experience, with a correct approach, yes it is. At least for acid/base equilibria. See equation 11.16 - while it is not in a polynomial form, converting it to one is just an exercise in patience.

Quote
After that, substitution of numerical values into the extra terms (Ca, Ka, Kw, etc.) yields an equation that can always be solved so long as there is a real solution.

Numerically, yes.
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Offline Big-Daddy

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Re: simultaneous equilibria
« Reply #21 on: November 01, 2012, 11:02:10 AM »
Strong acid in water yields a second degree equation in [H+]. Each additional dissociation constant increases the degree by 1, so solution for a monoprotic weak acid will be a polynomial of a 3rd degree, of diprotic acid 4th degree, of triprotic acid - 5th degree, and it is a known fact that there is no analytical solution for 5th degree polynomials.

Yes, but to find the polynomial in [H+]? That should always be possible, no?

In my experience, with a correct approach, yes it is. At least for acid/base equilibria. See equation 11.16 - while it is not in a polynomial form, converting it to one is just an exercise in patience.

Quote
After that, substitution of numerical values into the extra terms (Ca, Ka, Kw, etc.) yields an equation that can always be solved so long as there is a real solution.

Numerically, yes.

I can't find the necessary polynomial in the case of this problem. But I suppose this one doesn't involve common equilibrium constant terms like Ka, Kw, Ksp, etc. so could be impossible to solve.

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