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Topic: Equilibrium calculation of Mg  (Read 332 times)

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Offline LaGÅm

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Equilibrium calculation of Mg
« on: April 01, 2021, 02:34:34 PM »
The problem is as follows:
Mg2+ +CO32-  ::equil:: MgCO3
Using the K for the equilibrium, Kf=102.88 and the known concentration of MG species CMg=53.0 mM one should make an expression for [MG2+]

[MgCO3]/[Mg2+][CO32-]=Kf

from this i get
[Mg2+]=[MgCO3]/[CO32-]Kf

I have tried to attack this problem from diffrent angles but still don't arrive at the book solution of:
CMg*1/(1+[CO32-]Kf)
Help and ideas would be reatly appreciated


Offline Orcio_Dojek

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Re: Equilibrium calculation of Mg
« Reply #1 on: April 01, 2021, 04:23:14 PM »
Given concentration (53 mmol / liter) is far above solubility of MgCO3 (1,648 mmol / liter).

Maybe it is the sum of dissolved MgCO3 and MgCO3 as suspension.

Offline mjc123

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Re: Equilibrium calculation of Mg
« Reply #2 on: April 01, 2021, 04:47:49 PM »
If you assume CMg = [Mg2+] + [MgCO3] you can get the quoted answer. However, I have my doubts about this. I doubt if there is any significant concentration of undissociated MgCO3 in water. If it is present as the solid (even as a suspension) its activity will be 1.

Offline Orcio_Dojek

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Re: Equilibrium calculation of Mg
« Reply #3 on: April 01, 2021, 05:24:06 PM »
@LaGÅm

K = [Mg2+] [CO32-] / [MgCO3]

and [Mg2+] = [CO32-]

and [MgCO3] = C - [Mg2+]

so one way or another it still leads to quadratic equation.
« Last Edit: April 01, 2021, 06:22:16 PM by Orcio_Dojek »

Offline Orcio_Dojek

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Re: Equilibrium calculation of Mg
« Reply #4 on: April 02, 2021, 05:17:14 AM »
@mjc123

I wonder how did you get "correct" answer.

In my opinion:

[Mg]2 + K[Mg] = K C

and

K[Mg]2 + [Mg] = C       (answer given by author)

are not the same.

Offline mjc123

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Re: Equilibrium calculation of Mg
« Reply #5 on: April 02, 2021, 05:54:53 AM »
CMg = [Mg2+] + [MgCO3]
          = [Mg2+] + [Mg2+][CO32-]Kf
          = [Mg2+](1 + [CO32-]Kf)
[Mg2+] = CMg/(1 + [CO32-]Kf)

Compare the definitions of Kf and your K.

Note that this doesn't assume that [Mg2+] = [CO32-].

Offline LaGÅm

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Re: Equilibrium calculation of Mg
« Reply #6 on: April 02, 2021, 06:08:25 AM »

I just solved it the same way as above, my mistake was expanding both [Mg2+] and [MgCO3] when only [MgCO3] needed to be expanded!

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