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Offline mx4ly

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Equilibria and Mass Balance
« on: February 06, 2012, 08:36:49 PM »
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

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #1 on: February 07, 2012, 05:01:17 AM »
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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #2 on: February 07, 2012, 09:56:24 AM »
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. 


Offline Borek

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Re: Equilibria and Mass Balance
« Reply #3 on: February 07, 2012, 11:25:07 AM »
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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #4 on: February 07, 2012, 06:51:03 PM »
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.

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #5 on: February 08, 2012, 02:42:48 PM »
[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.

Quote
[ 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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #6 on: February 08, 2012, 05:13:47 PM »
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 

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #7 on: February 08, 2012, 05:30:54 PM »
            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?

Quote
[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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #8 on: February 08, 2012, 05:46:47 PM »
            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?

Quote
[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?

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #9 on: February 08, 2012, 06:29:36 PM »
[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.

Quote
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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #10 on: February 08, 2012, 07:56:06 PM »
Quote
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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Re: Equilibria and Mass Balance
« Reply #11 on: February 08, 2012, 09:42:16 PM »
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?

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #12 on: February 09, 2012, 04:09:21 AM »
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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Offline mx4ly

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Re: Equilibria and Mass Balance
« Reply #13 on: February 09, 2012, 08:08:28 AM »
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?

 

Offline Borek

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Re: Equilibria and Mass Balance
« Reply #14 on: February 09, 2012, 09:23:46 AM »
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 - 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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