[baboom]

Hello,

In my lab, we had to determine the concentration of Selenium in the shampoo.
The following equation summarizes what we did in the lab:

3Se+6(Ag+)+6NH3+3H2O-----> 2Ag2Se (s) + Ag2SeO3 (s)

We added Ag+ in excess, acidified the solution, and performed a back titration using KSCN to determine the amount of Ag+ in excess. The ferric indicator was added. When the excess Ag+ has completely reacted with the SCN- , the ferric indicator in the solution starts to react with the SCN- ion to make the red complex [Fe(SCN)]2+ . This means that when the solution starts to turn red, all the excess Ag+ has reacted with the KSCN.
 (Ag+)+(SCN-)--->AgSCN
(Fe3+)+ (SCN-)----> [Fe(SCN)]2+
My question is: since Ag2Se  is insoluble in acid, and Ag2SeO3 is soluble in acid, how does that affect the analysis?


I would really appreciate the help, Thanks

[baboom]

Wouldn't the Ag2Se and the Ferric indicator create some sort of a coloured complex which would affect the end point reading?

[baboom]

Or maybe the fact that Ag2SeO3 is soluble in acid and therefore dissolved in the solution, the SCN- ions will replace the SeO3-2 ions in the solution, and create a higher end point?

[Borek]

Or maybe the fact that Ag2SeO3 is soluble in acid and therefore dissolved in the solution, the SCN- ions will replace the SeO3-2 ions in the solution, and create a higher end point?

If Ag₂SeO₃ dissolves, concentration of Ag⁺ is higher than expected - not only excess is present in the solution, but also some of the ions expected to be part of the precipitate.

[baboom]

Hi Borek,
Thanks for your reply.
So you mean that the SeO3-2 ions will also be present in the solution and will affect the solubility of the AgSCN?

[baboom]

Ok, so how can I calculate the concentration of Se if my end point is not exact?

[Borek]

To some extent for sure.

[baboom]

Thanks Borek,
Much appreciated 

[Borek]

Ok, so how can I calculate the concentration of Se if my end point is not exact?

For exact calculation you would need pKa1 and pKa2 as well as exact pH value. But I would start ignoring the problem.