# Redox titration end point calculation -- Help please

1. Oct 1, 2014

### kirsten_2009

1. The problem statement, all variables and given/known data

A student weighs by difference 0.1956 g of sodium oxalate into a 100 ml Volumetric flask and dilutes to the mark 10 ml of HCl and 90 ml of distilled water. Approximately how many ml's of 0.02 M KMnO4 would be required to reach the equivalence point of the titration?

2. Relevant equations

3. The attempt at a solution

Please correct me if my reasoning is faulty. So, isn't the "end point" the point during the reaction in which the moles of sodium oxalate and potassium permanganate equal? And if this is so...then:

0.1956 g Na2C2O4 x 1 mol Na2C2O4 / 133.998 g/mol Na2C2O4 = 0.00146 mol Na2C2O4/0.1 L = 0.0146 M

and then there is 0.02 mol KMnO4/L

But I can't make the connection on how to find the information I'm asked....help? Thanks in advance!

2. Oct 2, 2014

### Staff: Mentor

No, it is not about equal number of moles, it is about stoichiometric number of moles.

See if that helps:

http://www.titrations.info/titration-calculation

3. Oct 2, 2014

### SteamKing

Staff Emeritus

4. Oct 6, 2014

### kirsten_2009

Hello,

Thank you for replying. I think I understand and the link you posted was very helpful :) so...would this be correct?

Taking into account the following redox reaction:

Reduction: MnO4-(aq) + 8H+(aq) +5e- >>> Mn2+(aq) + 4H20(l)
Oxidation: C2O42-(aq) >>> 2CO2(g) +2e-
Combining and Balancing: 2MnO42-(aq) + 5C2O42-(aq) + 16H+(aq) >>> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

Step #1: If I'm given the grams of sodium acetate as 0.1956 I can find moles:

0.1956 g Na2(C2O4) x 1 mol/134 g/mol = 0.001459 mol Na2(C2O4)

Step #2: Then I can find concentration:

0.001459 moles of solute / 100 mL of solution = 0.001459 / 0.1 L = 0.01459 M

Step #3: Then, using the balanced equation and stoichiometric ratios, I can find the moles of MnO42- :

0.01459 mol Na2(C2O4) x 2 mol MnO42- / 5 mol C2O4 = 0.0058388 mol MnO42-

Step #4: Then using the molarity of MnO42- , I can find the mL's needed:

0.0058388 mol MnO42- x 1 L of solution / 0.02 mol KMnO4- = 0.2919 L or 291.9 mL

5. Oct 6, 2014

### Staff: Mentor

Which you don't need, and which in the end made you make a mistake - you assumed you titrated not the original amount of oxalate, but amount present in a liter.

Other than that you are right, you just confused yourself and your final result is ten times off.

6. Oct 6, 2014

Thank You!