PhreeqcUsers Discussion Forum

Welcome Guest
 

  • Forum Home
  • Login
  • Register

  • PhreeqcUsers Discussion Forum »
  • Conceptual Models »
  • Equilibrium assumptions »
  • S(-2) speciation
« previous next »
  • Print
Pages: [1]   Go Down

Author Topic: S(-2) speciation  (Read 14125 times)

Szasza

  • Contributor
  • Posts: 1
S(-2) speciation
« on: 12/08/15 13:37 »
Hi,

I need some helps.

I made equilibrium model, and I calculate (from molality) the individual speciations how many percent from the total.
For example:
Ni [mol/L] = 6,95E-08            
Ni+2 [mol/L] =  4,46E-08  --> Ni2+ [%] = 64,2
NiSO4 [mol/L]  2,49E-08  --> NiSO4 [%] = 35,8
It gives 100%.

But the problem is with S(-2).
S(-2) [mol/L] = 1,11E-02
S5−2 [mol/L] = 2,22E-03 --> S5−2 [%] = 20,00
S4-2 [mol/L] = 1,04E-11 --> S42- [%] = 9,32E-08
This never gives 100%. Why? Where is the other 80%?
 S(6) speciations give 100%...

Could somebody help me?

Thank you!

Best regards,
Szasza

Logged

dlparkhurst

  • Global Moderator
  • *****
  • Posts: 4338
Re: S(-2) speciation
« Reply #1 on: 12/08/15 13:42 »
I suspect you are not accounting for the stoichiometry of the species. There are 5 moles of S in S5-2 (not that I am much of a believer in the polysulfide species anyway).
Logged

  • Print
Pages: [1]   Go Up
« previous next »
  • PhreeqcUsers Discussion Forum »
  • Conceptual Models »
  • Equilibrium assumptions »
  • S(-2) speciation
 

  • SMF 2.0.19 | SMF © 2021, Simple Machines | Terms and Policies
  • XHTML
  • RSS
  • WAP2