[mx4ly]

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.

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.

[Borek]

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).

[mx4ly]

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

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?

[Borek]

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)₄^2-, 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.

[mx4ly]

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)₄^2-, 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. 

[Borek]

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

Mn_initial = 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 K_sp and pH, after that all we can say is that concentration of Mn²⁺ is lower than ## K_{sp}/[OH^-]^2 ##.

[mx4ly]

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

Mn_initial = 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 K_sp 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.

[Borek]

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.

[mx4ly]

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. 

[Borek]

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.

[mx4ly]

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?   

[Borek]

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).

[mx4ly]

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?

[Borek]

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?

[mx4ly]

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.