Naked Science Forum

Non Life Sciences => Chemistry => Topic started by: frethack on 27/03/2009 05:02:06

Title: Titration question
Post by: frethack on 27/03/2009 05:02:06

Sorry to ask this guys...its for my homework in my chem lab

We are doing redox titration and one of our post-lab questions is this

Quote

Calculate the molarity of a FeSO4 solution if 80.0 ml of this solution requires 74.8 ml of 45.2 mM Na2Cr2O7 for the reaction to go to completion.

I have to write balanced equation and balanced half reactions as well.  I cant for the life of me figure out the balanced equation because I am drawing a blank on what the products would be. Aside from that, Im golden (I think...hehehe...Im an A student in chem...I promise!)

Title: Titration question
Post by: Chemistry4me on 27/03/2009 05:11:20

The oxidation half-equation is:

Fe²⁺ → Fe³⁺ + e^-

The reduction half-equation is:

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Overall: Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Can you do the rest now?  [:)]

Title: Titration question
Post by: frethack on 27/03/2009 05:17:08

Thats almost exactly what I have!!!...This whole time its been driving me nuts. The only difference is that I put Na2SO4 as a product as well.  Do I not have to account for the sulfate ion and the sodium?

Thank you very much, btw  [;D]  I think I can handle it after this  *hides head in embarassment*

Title: Titration question
Post by: Chemistry4me on 27/03/2009 05:20:48

Quote from: frethack on 27/03/2009 05:17:08

Do I not have to account for the sulfate ion and the sodium?

Usually if you don't include those spectator ions, it does not matter, it also makes things less complicated. [:)]

Title: Titration question
Post by: frethack on 27/03/2009 05:31:36

Thanks!