alkaline solutions for titration

[mgsephton]

Hi

I'm interested in simulating treatment of acid mine drainage with sodium hydroxide.  So far I have learned three ways (listed below) of defining a sodium hydroxide solution (0.05 M for example).

I am wondering if these are truly equivalent or whether I should avoid one or more of these approaches for any reason?  And are there any better or other ways I should consider using to define sodium hydroxide solution for the purpose of simulating treatment of acid mine drainage?
Many thanks
Michael


SOLUTION 1   0.05M NaOH
-water 1.0 #kg
EQUILIBRIUM_PHASES
NaOH   1.0   0.05 # moles
END

SOLUTION 2   0.05M NaOH
pH 7.0 charge
units mmol/L
Na   50
END

SOLUTION 3 0.05M NaOH
-water 1.0 kg
REACTION 1
   NaOH 1.0
   0.05 mole
END

The first way requires defining NaOH as a phase using:
PHASES
NaOH
1.0000 Na+ + 1.0000 H2O  =  NaOH +1.0000 H+
        log_k           -14.7948
   -delta_H   53.6514   kJ/mol   # Calculated

enthalpy of reaction   NaOH
#   Enthalpy of formation:   -112.927 kcal/mol
        -analytic 8.7326e+001 2.3555e-002 -5.4770e

+003 -3.6678e+001 -8.5489e+001
#       -Range:  0-300

(copied from the llnl.dat database)

[dlparkhurst]

There are two different solutions defined. Method 1 and 3 create 0.05 mol/kgw, whereas method 2 is 0.05 mol/L, which results in a slightly different concentration in terms of mol/kgw. The difference will not be very much.

I would not use the first method, just because it is a bit misleading. EQUILIBRIUM_PHASES is used, but in fact the key is the amount of NaOH provided (0.05 moles), and the equilibrium constant is really irrelevant because the solubility is limited by the amount available.

Method 2 and 3 are equally appropriate, although the concentrations are slightly different.

Depending on the volume of titrant, it may not be much different just to add NaOH to an acidic solution directly with REACTION. It is simpler to add multiple amounts of a REACTION than to set up an equal number of MIXes.