Question about setting Exchange Master Species?

[Jeonghwan Hwang]

Hello, my name is Jeonghwan Hwang,
Thank you for reading my article first.

When selecting an Exchange master species, can I set the initial state to AK, but not to A-?
I would like to express the model of cesium adsorption on illite under DIW using PHREEQC's ion exchange model.
Set the initial solution to no ions, set the master species and run PHREEQC,
Initially all master species can be seen to exist in the form of HA.
However, since the interlayer cation of illite is potassium, there is an error in the initial state expressed as HA.
Is there a way to initially set HA to KA, except for setting potassium in the initial solution?

Thank you for your help,
Sincerely

Jeonghwan Hwang

[dlparkhurst]

I think this is what you described (using X-, instead of A-). The initial exchange composition is defined as KX.

Code: [Select]

SOLUTION
EXCHANGE 1
    KX      1
END


[Jeonghwan Hwang]

Hello, thank you for your advice,
I tried to exchange initial Z, D, X to KZ, KD, KX.
However, there were not different results from before which used Z, D, X.
Can you help me to find my fault?
My code is shown as below;


 Title Initial solution
 Solution 1
    pH 7
    Units moles/l
    temp    25.0
 save solution 1
 end
 
 TITLE Caesium Sorption experiment
 use solution 1
 
 EXCHANGE_MASTER_SPECIES
    Z   Z- #FES
    D   D- #TypeII
    X   X- #planar

use solution 1
 
 EXCHANGE_MASTER_SPECIES
    Z   Z- #FES
    D   D- #TypeII
    X   X- #planar
 
 EXCHANGE_SPECIES
    Z- = Z- #FES
    log_k 0.0
    H+ + Z- = HZ #FES
    log_k 0.0
    Na+ + Z- = NaZ #FES
    log_k -1.80
    K+ + Z- = KZ #FES
    log_k 0.36
    Ca+2 + 2Z- = CaZ2 #FES
    log_k -10.28
 
    D- = D- #TypeII
    log_k 0.0
    H+ + D- = HD #TypeII
    log_k 0.0
    Na+ + D- = NaD #TypeII
    log_k -3.60
    K+ + D- = KD #TypeII
    log_k -1.47
    Ca+2 + 2D- = CaD2 #TypeII
    log_k -6.09
 
    X- = X- #planar
    log_k 0.0
    H+ + X- = HX #planar
    log_k 0.0
    Na+ + X- = NaX #planar
    log_k -2.50
    K+ + X- = KX #planar
    log_k -1.51
    Ca+2 + 2X- = CaX2 #planar
    log_k -5.37

 EXCHANGE 1
      KZ  1.65E-07
      KD  6.00E-06
      KX  2.38E-05
 -equilibrate solution 1
 end
 
Thank you!

Sincerely,

Jeonghwan Hwang

[dlparkhurst]

Remove the following line:

-equilibrate solution 1

You want to set the initial condition to be KZ, KD, and KX. By setting equilibrium with solution 1, H+ is the only cation, which forces the initial exchange composition to be HZ, HD, and KX.

[Jeonghwan Hwang]

Thank you for help,
I solved my problem.
I will try to make good research results.

Sincerely,

Jeonghwan Hwang