Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: mx4ly on February 06, 2012, 08:36:49 PM
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Hi guys,
I've been struggling with a fairly challenging (for me!) chemistry problem for about a week and a half now. I was hoping someone might be able to help me. Please let me know I what need to clarify, I am paraphrasing the problem here. I will try to state it as simply as I can without leaving anything out.
Manganese ions react with ammonia in the following reaction:
Mn2+ + NH3 <-> [M(NH3)]2+ log K = 1.00
However it can react with multiple ammonia molecules.
Mn2+ + 2NH3 <-> [M(NH3)2]2+ log K = 1.54
Mn2+ + 3NH3 <-> [M(NH3)3]2+ log K = 1.70
Mn2+ + 4NH3 <-> [M(NH3)4]2+ log K = 1.3
I am trying to determine the total concentration of Mn(NH3)n 2+ (M) at varying pHs.
Manganese associates with hydroxide ions in the following reaction:
Mn(OH)2 <-> Mn2+ + 2OH- Ksp = -12.70
Ammonia also associates with water in the following reaction. I am not sure how this fits into the problem at all.
NH3 + H20 <-> NH4+ + OH- Kb = -4.80
The mass balances are given as:
C(Mn2+) = [Mn2+] + [Mn(OH)2] + [Mn(NH3)2 2+] + [Mn(NH3)3 2+] + [Mn(NH3)4 2+]
C(NH3) = [NH3] + [Mn(NH3) 2+] + 2[Mn(NH3)2 2+] + 3[Mn(NH3)3 2+] + 4[Mn(NH3)4 2+]
Again, I am not sure how these fit into the problem
Additionally I am given:
Mn(OH)2 + nNH3 <-> [Mn(NH3)n] 2+ + 2OH-
C(Mn2+) = 2M
C(NH3) = 1M
Attempt at solution:
Basically, I am being a ton of information, and am not sure how to use it all. I am trying to determine how varying the pH will effect the resulting concentration of [Mn(NH3)i 2+].
First, we need to establish a relation between pH and [Mn2+]. I begin with
Ksp = [Mn 2+] * [OH-]^2
But already I am confused. We are given a Ksp value to use to solve the equation, yet [Mn 2+] also behaves as equaling [OH-]/2. Assuming we fix [OH-] (the pH), I am not sure which to use to calculate [Mn 2+].
Additionally I have
K = [Mn(NH3)]2+ / ([M2+]*[NH3])
Yet solving for the initial concentration of Mn2+ = 2M and [NH3] = 1M yields [Mn(NH3)]2+ = 5 M. I know this also does not relate the concentration of [Mn(NH3)]2+ to pH, so clearly I'm doing something wrong in this step.
I have a graph of the results that I am trying to replicate, which shows that [Mn(NH3)n]2+ to be about 0.2 at pH = 8. Any help on this would be greatly appreciated. I'm not expecting anyone to solve the entire problem for me - mainly I would like to know how to correlate the pH of the solution with the other concentrations I am trying to calculate.
Thanks (: Any advice on what I am doing wrong would be appreciated
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yet [Mn 2+] also behaves as equaling [OH-]/2
No. Mn2+ reacts further, and - if pH is kept constant - OH- is also consumed by some other process. So the concentrations are not related by the stoichiometry of dissolution, but - as long as there is a solid Mn(OH)2 present - only by Ksp.
Can you describe the experiment? It is not clear to me what is the setup - are you mixing solutions, or do you start with a solid hydroxide dissolved in ammonia solution, and how do you know the pH?
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Thanks for your reply.
The purpose of the experiment is to prepare spherical manganese hydroxide particles and analyze the effect that changes in pH have on their formation. You begin with an ammonia solution, and I believe you set the pH with the concentration of hydroxide ions. Manganese then reacts with hydroxide ions and ammonia to yield Mn(OH)2 and Mn(NH3)n 2+.
Here is the experiment discussed in more detail (the experiment is discussed above where table 2 is located): http://pubs.acs.org/doi/full/10.1021/cm803144d
I am trying to replicate the figure 6 graph which shows the concentration of Mn(NH3)n 2+ with the respect to the pH of the solution.
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I don't have access to the paper, but I am starting to feel what it is all about.
Standard approach is to write all equations - that means all equilibrium constants, all mass balances and charge balance - and try to solve them. It is never easy. But there is no better starting point.
If you know pH, you can calculate maximum possible concentration of free Mn2+ (from Ksp). That simplifies the problem, although there is still plenty of unknowns.
If there is a large enough excess of ammonia, you can assume [NH3]+[NH4+] equals initial concentration of ammonia - that would mean you can calculate [NH3] from Kb and pH. That simplifies the problem further.
Once you know these two values you should be able to calculate concentrations of all complexes, using overall stability constants.
If the excess of the ammonia is not large enough, it gets more complicated.
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Great response, thanks. Would you please be able to confirm if I am doing this right?
Let's assume a pH of 8
Mn(OH)2 <--> Mn2+ + 2OH-
Ksp = [Mn 2+] [OH-]^2
[Mn 2+] = 10^(-12.7) / ( 10^(-(14-8)) )^2
[Mn 2+] = .1995 at pH 8.
I'm not sure if this is right, because at low pH values, [Mn 2+] spikes up to ridiculous values (~10^11 at pH 2). Could you please explain what I'm doing wrong, or why this is the case?
Let's assume an excess of ammonia. We have:
NH3 + H20 <--> NH4+ + OH-
pH = 8 so [OH]- final = 10^(-(14-8)) = 10^(-6)
[ NH4 ] = [OH-] = 10^-6
x = 10^-6
[NH3] initial i = i - 10^-6
Kb = x^2 / ( i - x)
10^(-4.80) = (10^-6)^2 / (i - 10^-6)
[NH3]initial = i = 1.063 * 10^-6
This seems to be rather low to me. I'm almost certain I've done something wrong in this step, and probably the first one as well.
Once we have calculated the two values above, do we just plug the resulting Mn2+ and NH3 concentrations into each equilibrium reaction to determine the amount of [Mn(NH3)2+] produced?
Thanks again for your help so far, I really appreciate it.
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[Mn 2+] = .1995 at pH 8.
I'm not sure if this is right, because at low pH values, [Mn 2+] spikes up to ridiculous values (~10^11 at pH 2). Could you please explain what I'm doing wrong, or why this is the case?
At low pH there will be no solid present, so Ksp doesn't matter. Remember precipitate appears only when the product of concentrations (in right powers) is higher than Ksp.
[ NH4 ] = [OH-]
No. That will work only if reaction with ammonia is the only source of OH-. pH was adjusted by some other means, so these concentrations don't have to be identical.
To calculate concentration of free ammonia you need pH and information about the analytical (total) initial concentration of the ammonia. I see what I wrote yesterday was a little bit ambiguous - you need Kb, pH AND total concentration, mentioned in the same phrase earlier.
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Hmmm. I still think I'm missing something, but here is take II
The initial concentration of free ammonia was specified to be 1M in the original problem (and free Mn 2+ to be 2M)
Again assuming pH 8
NH3 + H20 <--> NH4+ + OH-
I 1 - 0 10^-6
C -x - +x +x
E (1-x) - x 10^-6 + x
Kb = 10^-4.8 = (10^-6 + x) * x / (1-x)
x = .00397
So the [NH3] = 1 - .00397 = .99603
Hardly a significant change; I assume something is wrong. I've double checked everything but it seems to be consistent with what I have learned in gen chem and what you've told me.
Anyways, proceeding to the next part of the problem:
Mn 2+ + NH3 <---> [Mn(NH3)]2+ log K = 1.00
[Mn 2+] = 2 - .1995 = 1.8005
[NH3] = .99603
K = 10^1 = [Mn(NH3) 2+ ] / ([Mn 2+ ] * [NH3])
[Mn(NH3) 2+] = 17.9335 - a ridiculously high concentration
As I mentioned before - the concentration [Mn(NH3)n 2+] at pH 8 appears is around 0.2. Performing a equilibrium balance around only 1 reaction grossly overshoots this. I've tried searching online for resources to help me with this problem but wasn't really able to find anything aside from standard general chemistry guidelines. Do you see the error I am making in this problem? Thanks again
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NH3 + H20 <--> NH4+ + OH-
I 1 - 0 10^-6
C -x - +x +x
This is (almost) equivalent to assuming [NH4+] = [OH-] - which is a wrong assumption.
Take a look:
$$ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} $$
$$ [NH_4^+]+[NH_3]=1 $$
What is known, what is unknown?
[Mn(NH3) 2+] = 17.9335 - a ridiculously high concentration
When dealing with approximate equilibrium concentration it is always important to check if the final results look reasonable. If they don't, it usually means either that the input data is wrong, or some of the assumptions are wrong. At the moment we already know concentration of the free ammonia is incorrect.
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NH3 + H20 <--> NH4+ + OH-
I 1 - 0 10^-6
C -x - +x +x
This is (almost) equivalent to assuming [NH4+] = [OH-] - which is a wrong assumption.
Take a look:
$$ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} $$
$$ [NH_4^+]+[NH_3]=1 $$
What is known, what is unknown?
[Mn(NH3) 2+] = 17.9335 - a ridiculously high concentration
When dealing with approximate equilibrium concentration it is always important to check if the final results look reasonable. If they don't, it usually means either that the input data is wrong, or some of the assumptions are wrong. At the moment we already know concentration of the free ammonia is incorrect.
Kb is known (given as 10^-4.8)
[OH-] is known (given as 10^-6) However, isn't this the initial pH? Won't more OH- form when ammonia reacts with water?
The initial concentration of NH3 is known (1M)
The initial concentration of NH4+ is known (0)
The final concentrations of NH3 and NH4+ are unknown. However, we know that the sum of the concentrations must equal 1.
So, from what you have written:
[NH4+] = 1 - [NH3]
But if we say x = [NH3]
Then we have:
Kb = (1-x)(10^-6)/(x)
Which is essentially what I just found to be incorrect :/ Could I get one more hint on what I'm missing?
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[OH-] is known (given as 10^-6) However, isn't this the initial pH? Won't more OH- form when ammonia reacts with water?
No. You know pH is 8 AFTER the equilibrium has been reached.
Then we have:
Kb = (1-x)(10^-6)/(x)
Which is essentially what I just found to be incorrect :/
No, it gives completely different concentration of NH3.
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Then we have:
Kb = (1-x)(10^-6)/(x)
Which is essentially what I just found to be incorrect :/
No, it gives completely different concentration of NH3.
Don't know how I missed that -_-
OK, so solving for x, x = 0.0593 so [NH3] at equilibrium = .0593 at pH 8, which seems reasonable (:
Now to work through the reactions
Mn 2+ + NH3 <---> [Mn(NH3)]2+ log K = 1.00
Mn 2+ + 2NH3 <---> [Mn(NH3)2]2+ log K = 1.54
Mn 2+ + 3NH3 <---> [Mn(NH3)3]2+ log K = 1.70
Mn 2+ + 4NH3 <---> [Mn(NH3)4]2+ log K = 1.30
I crunched a few numbers, and the results I am getting are very, very close to being correct. However, the results still seem to be a bit high at the central peak of [Mn(NH3)n]2+.
Specifically:
@ pH 7.5 -> [Mn(NH3)n]2+ reaches a near maximum slightly greater than 0.2.
However, working this out:
[Mn 2+] = Ksp / [OH-]^2 = 10^(-12.7) / ( 10^(-(14-7.5)) )^2
[Mn 2+] = 1.9952
Kb = 10^(-4.80) = (1-x)(10^-(14-7.5))/(x)
x = [NH3] = .01956
Equilibrium Balances:
[Mn(NH3)]2+ = 10^1 * 1.99526 * 0.01956 = 0.3903
[Mn(NH3)2]2+ = 10^1.54 * 1.99526 * 0.01956^2 = 0.02648
[Mn(NH3)3]2+ = 10^1.70 * 1.99526 * 0.01956^3 = 7.486*10^-4
[Mn(NH3)4]2+ = 10^1.3 * 1.99526 * 0.01956^4 = 5.83*10^-6
[Mn(NH3)n]2+ = .4175
This is about twice what it should be. However, values farther away from the maximum concentration seem to be more accurate. Do all of my calculations seem fine, and is the figure I am looking at possibly flawed, or am I missing an important step? I've also noticed that the last reaction, for [Mn(NH3)4]2+, hardly seems to contribute to the total concentration at all. Is this normal, assuming that [Mn 2+] will react with a smaller number of ammonia molecules far more than with multiple molecules?
Thanks very much for your help. I've made much more progress in the last two days than in the previous week and a half.
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Hmmmm, I'm now fairly certain something is wrong. I ran the same procedure on some additional reactions (different free metals) and the total concentration is nowhere close to where it should be. Any ideas?
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Hmmmm, I'm now fairly certain something is wrong. I ran the same procedure on some additional reactions (different free metals) and the total concentration is nowhere close to where it should be. Any ideas?
I am afraid it means used approximation (the one about the sufficient excess of ammonia) is incorrect (in other words: assumption that the ammonia concentration is constant is not meet). But it in turn means you need to make much more thorough analysis of the system. Back to the drawing board.
You will have to solve system of nonlinear equations - not a thing you can do by hand.
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Hmmmm, I'm now fairly certain something is wrong. I ran the same procedure on some additional reactions (different free metals) and the total concentration is nowhere close to where it should be. Any ideas?
I am afraid it means used approximation (the one about the sufficient excess of ammonia) is incorrect (in other words: assumption that the ammonia concentration is constant is not meet). But it in turn means you need to make much more thorough analysis of the system. Back to the drawing board.
You will have to solve system of nonlinear equations - not a thing you can do by hand.
So all six equations (K1, K2, K3, K4, Kb, and Ksp) will have to be solved at once will have to be solved at once, using differential equations? I might be able to write a program to do so in Matlab - but I have no idea how to set the equations up.
Do you know if there are any resources online I could take a look at to get an idea of how to solve them?
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No need for differential equations, this will be just a system of equations. Matlab sounds like a good idea. This is not much different from the general method of pH calculation (http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation) - every equilibrium can be calculated this way.
This system is a little bit tricky, as number of equations depends on whether Mn(OH)2 is present or not (alternatively, you can assume solubility product in the form
$$ [Mn^{2+}][OH^-]^2 \leq K_{sp} $$
in which case you have a system of equations and one inequality).
1M ammonia solution has pH around 11, so if the pH is 8, system was most likely acidified. You will need to account for the acid anion (be it Cl- or something else) in the charge balance. At the same time concentrations of both H+ and OH- are known from pH, which means there are two unknowns less.
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No need for differential equations, this will be just a system of equations. Matlab sounds like a good idea. This is not much different from the general method of pH calculation (http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation) - every equilibrium can be calculated this way.
This system is a little bit tricky, as number of equations depends on whether Mn(OH)2 is present or not (alternatively, you can assume solubility product in the form
$$ [Mn^{2+}][OH^-]^2 \leq K_{sp} $$
in which case you have a system of equations and one inequality).
1M ammonia solution has pH around 11, so if the pH is 8, system was most likely acidified. You will need to account for the acid anion (be it Cl- or something else) in the charge balance. At the same time concentrations of both H+ and OH- are known from pH, which means there are two unknowns less.
Thanks for your continuing help
I'm still a little confused as to how the equations are set up. Do we just have the six from before? I'm not sure how we would solve them, or what we are assuming we do not know now because the ammonia excess was eliminated.
Also, there isn't a mention of an acid anion - is the species used irrelevant? Otherwise I feel it would have been given or alluded to.
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I'm still a little confused as to how the equations are set up. Do we just have the six from before? I'm not sure how we would solve them, or what we are assuming we do not know now because the ammonia excess was eliminated.
You need to add mass balances and charge balance as well as water ion product (although if pH is given you can assume H+ and OH- are known, which is equivalent of using Kw equation). That should give n equations in n unknowns.
Also, there isn't a mention of an acid anion - is the species used irrelevant? Otherwise I feel it would have been given or alluded to.
If it is inert and doesn't react with any of the species present in the solution, it will be present only in the charge balance.
Note you can't change pH by magic - you have to either acidify or alkalize the solution, and that means adding either acid (H+ and some counteranion) or base (OH- and some countercation).
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Hi again,
I've been continuing to work on the problem, taking a slightly different approach than before which still isn't coming out quite right. I think this may have been what you were trying to tell me to do before, but I was a little confused by the method you were outlining. Anyways, here is my what I've done; unfortunately I'm still having some issues:
Basically we start with the ammonia mass balance:
C(NH3) = 1 = [NH3] + [M(NH3) 2+] + 2[M(NH3)2 2+] + 3[M(NH3)3 2+] + 4[M(NH3)4 2+] + [NH4+]
First we solve for the Mn concentration, as before.
Ksp = [Mn 2+] [OH-]^2 ~ .1995 at pH 8
Next, we every one of the mass balance terms in terms of NH3
Kb = 10^(-4.80) = [NH4+] [OH -] / [NH3]
[NH4+] = [NH3] * 10^(-4.80) / [OH-]
And each of the K expressions:
K = [M(NH3)n 2+] / ([NH3]^n * [Mn 2+])
[M(NH3)n 2+] = K * [NH3]^n * .1995 (pH 8)
All of these expressions are substituted into the original mass balance, and then we solve for [NH3], since [Mn 2+] is known. Knowing [NH3] we can solve for [M(NH3)n 2+] for each of the reactions.
This approach is giving a frustratingly close answer, but it is still incorrect. I've scoured through my code for the last few days and believe that the error stems from an oversimplification of the [Mn 2+] calculation which is later used to solve for all of the complex concentrations. The free manganese concentration that we calculate assumes that there are no other equilibrium balances in the system, and so cannot be directly plugged into the equilibrium reactions that yield complexes (I think). Is this correct, and if so, how do I adjust the manganese concentration to account for this?
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Your approach looks correct to me. If the differences are small, perhaps you are using different stability constants?
There are at least three reactions of Mn2+ with OH- that can take place in this solution:
Mn2+ + OH- ↔ MnOH+, Kf = 2.5x103
Mn2+ + 4 OH- ↔ Mn(OH)42-, Kf = 5x107
2 Mn2+ + 3 OH- ↔ Mn2(OH)3+, Kf = 1.3x1018
Also, please note that this solution has a rather high ionic strength, which means calculated results are only an approximation of the real values.
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Your approach looks correct to me. If the differences are small, perhaps you are using different stability constants?
There are at least three reactions of Mn2+ with OH- that can take place in this solution:
Mn2+ + OH- ↔ MnOH+, Kf = 2.5x103
Mn2+ + 4 OH- ↔ Mn(OH)42-, Kf = 5x107
2 Mn2+ + 3 OH- ↔ Mn2(OH)3+, Kf = 1.3x1018
Also, please note that this solution has a rather high ionic strength, which means calculated results are only an approximation of the real values.
The main problem that I see is with the Manganese balance. At medium pHs, the free manganese concentration is around 2. However, there will still be some manganese-ammonia complex that forms. So at pHs around 6, [Mn 2+] equals 2, and [Mn(NH3)n 2+] > 0. Then I go back and solve for the manganese hydroxide balance and find it is less than 0, which is impossible.
2 = [Mn 2+] + [Mn(OH)2] + [Mn(NH3)n], solving for [Mn(OH)2] yields a negative concentration.
Because of this I know something about the way I am calculating the free manganese concentration is incorrect. I'll take a look at the three reactions you listed and see if I can work them into the balance.
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At medium pHs, the free manganese concentration is around 2.
I don't see where are getting it from. All you can say is that initially you put some amont of Mn into the solution, and from thsi moment on Mn balance is
Mninitial = Mn(OH)2(s) + Mn2+ + all possible forms of all complexes, with stoichiometric coefficients.
Amount of Mn(OH)2(s) can be zero if enough Mn is complexed. As long as this solid exists, concentration of Mn is governed by the Ksp and pH, after that all we can say is that concentration of Mn2+ is lower than ## K_{sp}/[OH^-]^2 ##.
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At medium pHs, the free manganese concentration is around 2.
I don't see where are getting it from. All you can say is that initially you put some amont of Mn into the solution, and from thsi moment on Mn balance is
Mninitial = Mn(OH)2(s) + Mn2+ + all possible forms of all complexes, with stoichiometric coefficients.
Amount of Mn(OH)2(s) can be zero if enough Mn is complexed. As long as this solid exists, concentration of Mn is governed by the Ksp and pH, after that all we can say is that concentration of Mn2+ is lower than $$ K_{sp}/[OH^-]^2 $$.
Sorry, let me explain myself a bit more.
From the mass balance you wrote out: Mn initial = 2M.
At pH 6: [Mn 2+] = Ksp / [OH-]^2 = 10^-12.70 / (10^(-(14-6)))^2 = 1995.26
Now this is where I think I'm making an incorrect assumption: Because the calculated amount of free manganese at pH 6 is greater than the initial amount of free manganese, at that pH, the actual amount of free manganese is 2M.
However, when we plug [Mn 2+] = 2 into the complex reactions at pH 6, we get that the total amount of complex is greater than 0.
This implies that:
Mn initial = 2 = 2 + [Mn(OH)2] + [Complexes] where the concentration of complex is greater than 0
so [Mn(OH)2] is less than 0.
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At pH 6: [Mn 2+] = Ksp / [OH-]^2 = 10^-12.70 / (10^(-(14-6)))^2 = 1995.26
pH 6, or 8?
At pH 8 [OH-] = 10-6, so concentration of Mn2+ in equilibrium with the solid is 0.1995 (or 0.2).
At pH 6 yes - you get 1995M as a concentration of Mn2+ in equilibrium with solid. It is impossible, highest molar concentrations are usually below 20, so the result you got means there is no solid present at this pH.
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At pH 6: [Mn 2+] = Ksp / [OH-]^2 = 10^-12.70 / (10^(-(14-6)))^2 = 1995.26
pH 6, or 8?
At pH 8 [OH-] = 10-6, so concentration of Mn2+ in equilibrium with the solid is 0.1995 (or 0.2).
At pH 6 yes - you get 1995M as a concentration of Mn2+ in equilibrium with solid. It is impossible, highest molar concentrations are usually below 20, so the result you got means there is no solid present at this pH.
When you say "no solid present," this only applies to the Mn(OH)2, correct? Meaning there is complex (albeit a small amount) at pH 6. My question is how there can be ANY complex that forms at all when the concentration of [Mn 2+] is 2. This would seem to violate the manganese mass balance, as the initial concentration is 2M, the free manganese is 2M, yet their is a nonzero amount of complex in the solution.
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My question is how there can be ANY complex that forms at all when the concentration of [Mn 2+] is 2. This would seem to violate the manganese mass balance, as the initial concentration is 2M, the free manganese is 2M, yet their is a nonzero amount of complex in the solution.
Initial concentration of complexes is zero. When they are created, concentration of free Mn2+ goes down (stoichiometrically), and mass balance is OK.
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My question is how there can be ANY complex that forms at all when the concentration of [Mn 2+] is 2. This would seem to violate the manganese mass balance, as the initial concentration is 2M, the free manganese is 2M, yet their is a nonzero amount of complex in the solution.
Initial concentration of complexes is zero. When they are created, concentration of free Mn2+ goes down (stoichiometrically), and mass balance is OK.
I assumed something like this would occur which leads to another question where I believe the slight inaccuracies I am having stems from. How do I calculate the concentrations of the complexes while simultaneously accounting for the lowering of free [Mn 2+]? For instance: currently I am calculating all of the complexes using the relationship [Mn(NH3)n] = K * [Mn 2+] * [NH3] ^ n, where [Mn 2+] is always set equal to 2. However, shouldn't the actual value of the concentration [Mn 2+] be decreasing as the complex uses up the free manganese? How do I equate the formulas for the complexes with [Mn 2+] while simultaneously lowering the free metal concentration?
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currently I am calculating all of the complexes using the relationship [Mn(NH3)n] = K * [Mn 2+] * [NH3] ^ n, where [Mn 2+] is always set equal to 2.
It won't work. You have to write a set of all equations, then solve them numerically. When there was the solid present situation was easier, as solid presence defined metal concentration (if the concentration was going down, hydroxide was dissolving). Now there is no source of Mn2+, so its concentration is not constant (but mass balance still holds, as usual).
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Here is the system of equations I came up with:
OH = 10^(-(14-8));
syms w x y z a o c m
eq1 = w/(a*(m-(w+x+y+z))) - 10^1;
eq2 = x/(a^2*(m-(w+x+y+z))) - 10^1.54;
eq3 = y/(a^3*(m-(w+x+y+z))) - 10^1.70;
eq4 = z/(a^4*(m-(w+x+y+z))) - 10^1.30;
eq5 = o * OH / a - 10^(-4.80);
eq6 = (m) * OH^2 - 10^(-12.70);
eq7 = (m-(w+x+y+z)) + c + w + x + y + z - 2;
eq8 = a + w + 2*x + 3*y + 4*z + o - 1;
Where w, x, y, and z are the concentration of the four complexes (in order), a is the concentration of free ammonia, o is the concentration of NH4+, c is the concentration of solid, and m is the concentration of free manganese. 8 equations, 8 variables
Essentially, whereas before in equations 1, 2, 3, 4, and 7 I simply used m - the concentration of free manganese, I replaced m with (m - (w+x+y+z)), the concentration of free manganese minus the concentration of the total amount of complex. However, this system of equations cannot be solved for, meaning the substitution I am using is wrong. Can you identify where the system of equations is incorrect?
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Sorry, it would be much easier if you post these equations in their standard forms. I am ready to help, but I am not ready to dig through the non-standard variables and equations translating them to something readable.
You have 8 equations and 8 unknowns, what is the problem with solving? Do you get some impossible numbers as answers?
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Sorry, it would be much easier if you post these equations in their standard forms. I am ready to help, but I am not ready to dig through the non-standard variables and equations translating them to something readable.
You have 8 equations and 8 unknowns, what is the problem with solving? Do you get some impossible numbers as answers?
OH = 10^(-(14-8));
10^1 = [Mn 2+]/( [NH3] * ([Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
10^1.54 = [Mn 2+]/( [NH3]^2 * ([Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
10^1.70 = [Mn 2+]/( [NH3]^3 * ( [Mn^2+] -([Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
10^1.30 = [Mn 2+]/( [NH3]^4 * ( [Mn^2+] -([Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
10^(-4.80) = [NH4+] * OH / [NH3]
10^(-12.70) = [Mn 2+] * OH^2
2 = ( [Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )) + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4];
1 = [NH3] + [Mn(NH3)] + 2*[Mn(NH3)2] + 3*[Mn(NH3)3] + 4*[Mn(NH3)4] + [NH4 +]
The 8 variables are: [Mn 2+], [NH3], [NH4 +], [Mn(OH)2], [Mn(NH3)], [Mn(NH3)2], [Mn(NH3)3], and [Mn(NH3)4]
The italicized bits are the ones I am concerned about. If I replace the italicized portions with only [Mn 2+], I get the same solution as solving the equation normally (which is slightly incorrect). I have tried to introduce subtracting the total amount of complex from the free manganese concentration to account for its decrease as it reacts to form complex.
The problem is that when I introduce the italicized bits, solving for the equations reduces the complex solutions [Mn(NH3)], [Mn(NH3)2], etc to functions of [Mn(NH3)4]. I assume this indicates the equations are not independent. It also means the way I have set up the equations is incorrect.
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10^1 = [Mn 2+]/( [NH3] * ([Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
[Mn2+]/(...) or [Mn(NH3)2+)/(...)? Or is it just a typo?
Even if it is a typo, I don't see how these equations reduce to "functions of [Mn(NH3)4]". Could be it is possible to eliminate other variables - that would be OK, as it would probably mean they can be calculated once you find the value of the [Mn(NH3)4]. Somehow I find it unlikely.
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10^1 = [Mn 2+]/( [NH3] * ([Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )))
[Mn2+]/(...) or [Mn(NH3)2+)/(...)? Or is it just a typo?
Even if it is a typo, I don't see how these equations reduce to "functions of [Mn(NH3)4]". Could be it is possible to eliminate other variables - that would be OK, as it would probably mean they can be calculated once you find the value of the [Mn(NH3)4]. Somehow I find it unlikely.
[Mn(NH3)n]. Yes, that was a typo. Sorry about that.
I can't work out analytically how it reduces down either, as Matlab is doing all of the equation solving. However when I solve the system with italicized bits replaced with [Mn 2+] it works out correctly to the previous (incorrect) method I was using. So the overall system is set up correctly, except for the modifications I am making to take into account that the complex formation uses up the [Mn 2+].
I can't seem to identify the error. Do you see what is wrong with the system of equations?
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10^(-12.70) = [Mn 2+] * OH^2
That would mean presence of solid, I think we already checked there will be no solid at pH 6.
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10^(-12.70) = [Mn 2+] * OH^2
That would mean presence of solid, I think we already checked there will be no solid at pH 6.
Ok, let's ignore that equation if we are doing the low pH case. We eliminate that equation and one variable because [Mn(OH)2] = 0. It's still does not work out in Matlab - meaning the system of equations is still incorrect when no solid is present
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2 = ( [Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )) + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
I think I know what have happened. This is incorrect, you are mixing formal and equilibrium concentrations. It should be
2 = [Mn2+] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
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2 = ( [Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )) + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
I think I know what have happened. This is incorrect, you are mixing formal and equilibrium concentrations. It should be
2 = [Mn2+] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
Hmmm, I don't really understand what you mean here. It looks like you are counting each of the complex molecules as double in terms of solving for the manganese concentration - but I don't understand why that would be. Essentially you have: 2 = [Mn 2+] + [Mn(OH)2] + 2*[complex]. It seems to be working though - would you mind explaining the logic behind it? I'll try to extrapolate now to cases where there is complex present.
Another plus - I found a better equation solver than Matlab's default. The problems I was having before stemmed from a few numbers getting too small for Matlab to keep track of. I'm able to find the correct solution now that values are actually being returned.
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Sorry, my mistake. They should be listed only once.
Note that in your original equation
2 = ( [Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )) + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
complexes concentrations cancel out, leaving just
2 = [Mn^2+] + [Mn(OH)2]
Equation as I wrote it was wrong, but was "relatively" correct - that is, it was correct in logical terms, just the coefficient had incorrect value - so it was working better than equation that was in no way related to the situation in the solution.
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Sorry, my mistake. They should be listed only once.
Note that in your original equation
2 = ( [Mn^2+] -( [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4] )) + [Mn(OH)2] + [Mn(NH3)] + [Mn(NH3)2] + [Mn(NH3)3] + [Mn(NH3)4]
complexes concentrations cancel out, leaving just
2 = [Mn^2+] + [Mn(OH)2]
Equation as I wrote it was wrong, but was "relatively" correct - that is, it was correct in logical terms, just the coefficient had incorrect value - so it was working better than equation that was in no way related to the situation in the solution.
Ok, that makes sense.
Now we get to the case with [Mn(OH)2] formation. This begins pH ~ 7.5 (a peak in [complex]). I am trying to use this additional equation, relating the free manganese and total complex to the solubility product. I assume this is wrong because the resulting values are not correct, yet this makes logical sense to me:
([Mn 2+] + [complex]) * [OH-]^2 = 10^-12.70
The complex resulting from this system of equations should equal the case where solid = 0 at pH 7.5, which is unfortunately not happening. Additionally, the concentration of complex flattens out much more quickly then when pH is increasing up to that point. The total amount of complex at pH 8 is ~.072 calculated this way. At pH 7 the concentration of complex is around .095, and at the peak pH 7.5 it is around .23 (using [Mn(OH)2] = 0). The concentrations should be equal equidistant from the peak concentration, which is not the case.
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For Mn(OH)2 presence of the complex doesn't matter. All that matters are equilibrium concentrations of Mn2+ and OH-. We have discussed it extensively before.