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need help fixing pH

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All correct except number 3.

3) You put "Halite -20 10" so that there would be a little Cl- available to make the HCl reactant. You made the target saturation index strongly negative so that only a little Cl- and Na+ could be there, just enough to provide the necessary Cl-. Since K is being used to balance charge, its concentration can be reduced in order to allow a little Na+ to be in solution.

Adding the Halite equilibrium phase allows NaOH to be added instead of HCl if necessary to arrive at the specified pH.

If HCl is needed, HCl is added through the Fix_H+ reaction, and a tiny bit of NaCl dissolves to put enough Na+ in solution to have an SI of -20 for Halite.

If NaOH is needed, then NaCl dissolves and HCl is removed, leaving Na+ and OH- in solution. Concentration of Cl wil be quite small to obtain the -20 SI.

NaCl + H2O = Na+ + OH- + HCl(l)

The (l) simply indicates it is removed from solution.

Got it! So these means the bases* are covered no matter which way the pH needs to be adjusted, and only a little Na+ or Cl- will end up in there to make that happen.

I will pay it forward by sharing all of your help with the graduate student in my department who first did the calculations in an attempt to help me, so that she might be able to do a better job with her own thesis calculations (in GWB, with a completely different application from what my group is doing). 

Best wishes,

*No pun intended.

Dear David,
Can you please explain why HCl is removed from the solution when NaCl hydrolysis takes place at fixing pH by  Halite -20 10 ?

First, let's assume you use Fix_H+ and HCl and you need to decrease the pH; acid--HCl--will simply be added to the solution. The concentration of Cl- will be significant (>1e-8), and assuming Na concentration is zero initially, NaCl will dissolve a tiny amount so that the product [Na+][Cl-] = 10^-20. [Na+] will be 1e-12 or smaller.

If the pH needs to increase and Cl is present in solution, HCl will be removed from the solution, which you could represent with the following reaction (again HCl(l) means it is removed from the solution):

--- Code: ---H2O + Cl- = HCl(l) + OH-

--- End code ---

Now, there can be a few problems with this reaction. First, it is hard to imagine physically removing HCl from solution. Moreover, the reaction is limited by the amount of Cl in solution; you can only take out as much Cl as is present initially, which may not be enough to adjust to the desired pH (PHREEQC will fail to converge). If there is no Cl (as in the example), it is not possible to increase the pH in this way at all. A way around the problem is to add the NaCl. It will allow adding NaOH when base is needed.

If you need to add NaOH to raise the pH, you can get the Na from NaCl. However, you do not want to add the Cl to solution, so the Fix_pH with HCl allows you to remove Cl. In general, you will need a significant amount of NaOH to fix the pH, so the Na concentration will be 1e-8 mol/kgw or greater, but with log K for NaCl of -20, the Cl- concentration will be 1e-12 or smaller.

--- Code: ---solution + NaCl + H2O = solution + Na+ + OH- + HCl(l)

--- End code ---

The dissolution of halite and removal of HCl don't make sense individually, but the net reaction is simply to add NaOH to raise the pH.

Dear David,
Thank you for the explanation!


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