# The Naked Scientists Forum

### Author Topic: Titration question  (Read 3006 times)

#### frethack

• Sr. Member
• Posts: 394
##### Titration question
« 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!)

#### Chemistry4me

• Neilep Level Member
• Posts: 7709
##### Titration question
« Reply #1 on: 27/03/2009 05:11:20 »
The oxidation half-equation is:

Fe2+ → Fe3+ + e-

The reduction half-equation is:

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

Overall: Cr2O72- + 14H+ + 6Fe2+ → 6Fe3+ + 2Cr3+ + 7H2O

Can you do the rest now?

#### frethack

• Sr. Member
• Posts: 394
##### Titration question
« Reply #2 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    I think I can handle it after this  *hides head in embarassment*

#### Chemistry4me

• Neilep Level Member
• Posts: 7709
##### Titration question
« Reply #3 on: 27/03/2009 05:20:48 »
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.

#### frethack

• Sr. Member
• Posts: 394
##### Titration question
« Reply #4 on: 27/03/2009 05:31:36 »
Thanks!

#### The Naked Scientists Forum

##### Titration question
« Reply #4 on: 27/03/2009 05:31:36 »