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R.R. Jha
R.R. Jha, Tutor
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Experience:  B.Tech
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It may be hard to answer my question because it is from a lab

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It may be hard to answer my question because it is from a lab report I did for chemistry. Hopefully you will be able to help me without me having to give you too much detail on all the data I acquired. Do you happen to know how to get the percent complexation of an ion? Basically in general terms if I ended up with FeSCN+2 how would I find out the percent of Iron(lll) ion still left?

I'm RRJha and I'd be glad to help. I think what you're referring to is reaction of Fe3+ with SCN-.

Fe3+(aq.) + SCN-(aq.) <=> Fe(SCN)2+

That being the case, I believe you'd have already found the concentration of Fe[SCN]2+. The balanced equation above shows that both reactants have 1-to-1 ratio with the product. So, one-mole of SCN- produces one mole of Fe(SCN)2+.
So, final concentration of thicyanate ion is [SCN-]final = [SCN-]initial - [Fe(SCN)]2+.
Similarly, final concentration of ferric ion is [Fe3+] final = [Fe3+]initial - [Fe(SCN)]2+.

Let me know if any question, or if your chemical reaction is different from what I presumed.

Thanks and regards
Customer: replied 3 years ago.

i believe the equation you have posted is correct but i think my professor wants the percentage of iron left in the FeSCN i am not sure if there is an equation for that

Which values or concentration you've already measured or calculated?
Customer: replied 3 years ago.

absorbance, initial concentrations of Fe and SCN, the equilibrium for Fe, SCN, and FeSCN using an ICE table.Also epsol i think that it how its dpelled which is the slope when referring to the beers law.


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Customer: replied 3 years ago.

yes that does answer my question thank you so much

Most welcome!