Redox titration end point

[circulus vitios]

---

Members don't see this ad.

Flask 2: 1.52 g FeSO4 + enough water to make 200 mL solution.
Flask 3: 3.04 g FeSO4 + enough water to make 100 mL solution.

20 mL sample from flask 2 required 17.5 mL Na2Cr2O7 solution of unknown concentration to reach the endpoint.
How much Na2Cr2O7 solution is required to reach the equivalence of a 20 mL sample from flask 3?

Answer is 70 mL because flask 3 [FeSO4] is four times flask 2 [FeSO4].

I thought equivalence point was when moles of acid equal moles of base or whatever the analogy is for redox titrations, so I figured the answer would be 35 mL because there are twice as male moles of FeSO4 in flask 3 than flask 2. Why are we looking at concentrations?

 

[sp808]

You don't consider moles because they make the solution but then take the same volume (20 ml) from each flask. If you multiply the molarity by the volume (which is the same in each case) it will be a factor of 4. You're not titrating the entire flask.

 

[circulus vitios]

You don't consider moles because they make the solution but then take the same volume (20 ml) from each flask. If you multiply the molarity by the volume (which is the same in each case) it will be a factor of 4. You're not titrating the entire flask.

Duh, thanks! 👍