Redox Titrations question

  • Thread starter loadsy
  • Start date
  • #1
57
0
Hey there, recently we just did a lab in Chemistry 101 on "Redox Titrations." I'm now working on the lab writeup but was hoping that someone may be able to help me with this question I am currently stuck on. It asks the following:

A sample of Au2(C2O4)3 is contaminated with Li2(C2O4). An acidified 0.2000g sample of this mixture requires 18.12mL of 0.02593mol/L KMnO4 to reach the equivalence point. In acidic solution, the oxalate ion is protonated and titrated with MnO4^- as in the laboratory experiment. Determine the mass percent of Li2(C2O4) in the sample.

Note: The metal ions are not oxidized by the permanganate ion. However, all of the oxalate bonded to the metals reacts completely with the permanganate ion.

Now I know you can start off the problem by finding the number of mols and then from that finding the molar mass of Li2(C2O4). But from there what do I do? Thanks a lot to anyone who can help me. :D
 

Answers and Replies

  • #2
GCT
Science Advisor
Homework Helper
1,728
0
You need to understand which reactions are going on, please specify these reactions, even more, use the stoichiometry of the equations to solve the problem.
 
  • #3
57
0
Hmmm well I know the reactions between the substances and understand what is taking place. It's just the steps in between I'm having trouble understanding because for a mass percent equation you need the substances in grams and multiply by 100 to determine it, and I'm not sure how to go about doing so. Thanks for the advice though.
 
  • #4
Borek
Mentor
28,476
2,873
You have to set up system of two simultaneous equations: one for sample mass (sum of masses of both substances), second for number of moles of oxalates (from titration).
 
  • #5
57
0
Ahhh that makes a little more sense, and then after finding the answers to both of those equations I'll set up, I'll be able to find the mass percent of the Li2(C2O4)3 sample. I think I kinda get it now. Thanks for the help guys. If I have any further questions regarding this problem, I'll come back and ask.
 
  • #6
57
0
Alright I've answered that last question, but I'm having problems on a fairly similar question. The question asks:

An unknown metal oxalate complex, M2(C2O4)3, reacts with KMnO4. The metal ion is not oxidized by KMnO4, however, all of the oxalate bonded to the metal reacts completely with KMnO4. Titration of an acidified 0.4556g sample of M2(C2O4)3 requires 28.42mL of 0.06049mol/L KMnO4 to reach the equivalence point. In acidic solution, the oxalate ion is protonated and titrated with MnO4^- as in the laboratory experiement. What is the metal M?

Now this question, like I said seems fairly identical to the other one. You start off by finding the number of mols of KMnO4 and from that, finding the number of mols of M2(C2O4)3, which I found to be 265.02g/mol. However, the question asks "What is the metal M?" I know you have to find the molar mass of what M actually is in the question, but how do I go about doing that? I've thought about doing it this way:

Since we know the ratio 2:3 I thought maybe finding the molar mass of (C2O4) and multiplying everything by 2 since it has a charge of 2-, and then dividing that answer by 3 since there is a charge of 3+. Does that make sense? Because after doing that I found the metal to be Nickel since it's closest in atomic mass. However, have I done this right or do I just need to go cry into my textbook for a minute and try again?
 
  • #7
Borek
Mentor
28,476
2,873
loadsy said:
finding the number of mols of M2(C2O4)3, which I found to be 265.02g/mol
I suppose you meant 'molar mass', not number of moles.

I know you have to find the molar mass of what M actually is in the question, but how do I go about doing that?
Assuming you have calculated molar mass of the oxalate correctly:

2*(molar mass of the metal) + 3*(molar mass of oxalate anion) = 265.02

Molar mass of oxalate is easy to calculate, that leaves you with one unknown.
 
  • #8
57
0
Hahaha yeah I did mean molar mass. I think it makes a little more sense, now. I had it is a 6:1 ratio and it came out completely wrong. 2:3 ratio is where it's at. :D
 

Related Threads on Redox Titrations question

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
9
Views
6K
Replies
6
Views
813
  • Last Post
Replies
3
Views
3K
Top