[luketapis]

Hi everyone. Today we made an experiment- Redox titration. The experiment went nice and smooth but I don’t know how to answer the question about concentration. Can anyone help me or give me a hint?

This is the experiment:

Preparation of a standard Ethanedioic Acid Solution
Calculate the mass of ethanedioic acid dehydrate required to make 100cm^3 of a 0.03M standard solution.Then weight out the required mass and make up the standard solution of the ethanedioic in a 100cm^3 volumetric flask.

Set up a burette and fill it with KMnO4 solution. Now transfer 10.0cm^3 of the standard ethanedioic solution into a 100cm^3 conical flask using a bulb pipette. Then, using a measuring cylinder, add 12.0 cm^3 of 2M sulphuric acid to the flask. Now gently heat this solution using a Bunsen burner.
After heating titrate this solution against KMnO4 in the burette until two concordant titres have been obtained. The end point of this titration is when a pink colour is obtained. Record the titres you obtained.
Calculate the concentration of the KMnO₄.

I did titration 3 times and there are my results:
1st: 11.5ml
2nd: 11.7ml
3rd: 11.7ml
=34.9/3=11.63ml

So the things I know:
Mass of ethanedioic acid dehydrate: H₂C₂O₄2H₂O = 124g
V=0.1l
C=0.03M
n = 0.1 x 0.03 = 0.003 x 124=0.372
This experiment is redox titration, so
Mn0₄⁻+8H⁺+5e⁻ → Mn2⁺ + 4H₂O   /x2      reduction
(COOH) ₂ → 2CO₂ + 2e⁻ + 2H⁺         /x5      oxidation
_________________________________________
2Mn0₄⁻+6H⁺+10e⁻+ 5(COOH) ₂ →2Mn2⁺ + 8H₂O + 10CO₂ + 10e⁻ 
But I do not know what to do next. Can anyone help me? I have never made redox titrarions so I have no idea how to calculate concentration of KMnO₄.     

 

[Dan]

1. Calculate the number of moles of oxalic acid consumed again - you have used an incorrect molar mass (see below).
2. Using the stoichiometry of your chemical equation (the 10 electrons on each side cancel out by the way), work out the number of moles of permanganate originally present.
3. Using the titre volume and your answer to #2, calculate concentration.

A couple of things to point out:

1. It is dihydrate, not dehydrate.
2. The molar mass is 126 g/mol, not 124g
3. The symbol for litre is a capital L.

[luketapis]

Is that mean that I do not have to take in a account that H2SO4 was used in this experiment?

Then:
2Mno4 react with 5(COOH)2

(COOH):
c=0.03
v=0.01
n=0.0003

Mno4:
c?
v=0.001163
n=0.0003x2=0.0006

c=0.0006/0.001163 = 0.516mL-1
Is that ok?

[Dan]

Is that mean that I do not have to take in a account that H2SO4 was used in this experiment?

It is likely to be in excess, I would begin the question by assuming permanganate is the limiting reagent - it can be checked afterwards.

Then:
2Mno4 react with 5(COOH)2

yes

(COOH):
c=0.03
v=0.01
n=0.0003

Mno4:
c?
v=0.001163
n=0.0003x2=0.0006

No. You have correctly identified that 2 mol of MnO₄^- react with 5 mol oxalic acid. Think about this for a minute, it means there is less MnO₄^- than oxalic acid doesn't it? So what you have calculated, that there is twice as much MnO₄^- as oxalic acid, cannot be sensible.

[luketapis]

I understand:
5(cooh)2: n=0.0003
1(cooh)2: n= 0.0003/5=0.00006

2 MnO4 =2x 0.00006= 0.00012
C=n/v
c= 0.00012/0.001163 =0.132mol L-1

Am I right?

[Dan]

I understand:
5(cooh)2: n=0.0003
1(cooh)2: n= 0.0003/5=0.00006

2 MnO4 =2x 0.00006= 0.00012
C=n/v
c= 0.00012/0.001163 =0.132mol L-1

Am I right?

Close, method is fine, only problem is that 11.63 mL is not equal to 0.001163 L

Note that what you have written: 0.00012/0.001163 is not equal to 0.132, it is equal to 0.103. Take care with your calculations.

[luketapis]

0.00012/0.001163 = 1.03 thats right.
but 11.63 cm^3 = 0.001163L

[Dan]

but 11.63 cm^3 = 0.001163L

Not true:

1 cm³ = 1 x 10^-3 L

Therefore:

11.63 cm³ = 11.63 x 10^-3 L = 0.01163 L

0.001163 L = 1.163 cm³

[luketapis]

thanks for your help.Now I understand the whole problem of this excercise