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Topic: S(-2) speciation (Read 14127 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
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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).
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