Checking exchanger reactions

[peterwadeuk]

Greetings good people,

A colleague wished to solve an exchange reaction with Na, K, Ca, Mg, H analytically.

I tried to persuade him to use PHREEQC.

So as to simplify the initial induction I stripped down the WATEQ4F database, as attached.

We ran the program, output attached, and received the following output which he performed calculations on to verify.

For the divalent ions:

Reaction   Ca + MgX2 = Mg + CaX2   
      
   Species   Activities
   Ca   9.980E-04
   Mg   6.698E-04
   CaX2   4.977E-04
   MgX2   2.107E-04

Rxn No.   Reaction   LogK
(1   Ca + 2X = CaX2   0.8
(2   Mg + 2X = MgX2   0.6
(1 – (2   Ca + MgX2 = Mg + CaX2   0.2
      
Calculated LogK   (Mg * CaX2) / (Ca * MgX2)   0.20
      
The calculated LogK is in agreement with the LogK calculated from the component exchange reactions.

For combination monovalent and divalent ions:      
      
   2Na + MgX2 = Mg + 2NaX   
      
   Species   Activities
   Mg   6.698E-04
   MgX2   2.107E-04
   Na   9.011E-04
   NaX   1.388E-05
      
Rxn No.   Reaction   LogK
(3   2Na + 2X = 2NaX   0
(4   Mg + 2X = MgX2   0.6
(3 – (4   2Na + MgX2 = Mg + 2NaX   -0.6
      
Calculated LogK   (Mg * NaX^2) / (Na^2 * MgX2)   -3.12

!! The calculated LogK is not in agreement with the LogK calculated from the component exchange reactions.

We are doing something wrong here which I believe has something to do with the definition of the exchange equilibrium constants.

Can someone advise where the problem may lie?

Thank you,
Peter
 

[dlparkhurst]

I calculate the following:

Species     log(activity)
Na           -3.045
Mg           -3.174
NaX         -2.034
MgX2       -0.551

Log(IAP) = la(Mg) + 2*la(NaX) - 2*la(Na) - la(MgX2) = -0.6
             

[peterwadeuk]

Thank you David.

I realise now that I should get the activities by explicitly requesting them e.g. SELECTED_OUTPUT 1 as opposed to trying to find them in the phreeqc output file.

All the best, Peter