A redox titration with kmno4- and h2o2 help

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    Redox Titration
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The discussion focuses on a redox titration involving KMnO4 and H2O2, with the user seeking help to calculate the average volume of permanganate ion used. They mention using 14.32 mL of 0.025 M KMnO4 and inquire if this average should be used for further calculations. Participants clarify that the average volume refers to the total KMnO4 used during titration and emphasize the importance of understanding molarity and moles in the calculations. The user successfully completes their calculations and analysis but seeks confirmation on how a contaminant affecting H2O2 would influence the percent composition. It is concluded that the presence of a contaminant would indeed decrease the measured percent of hydrogen peroxide.
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Homework Statement


Calculate the average volume of permangate ion used.

I'm given the average volume of kmno4 used.

Homework Equations



N=m/mm

The Attempt at a Solution



I tried to just average my kmno4 used, but I don't believe that's what my teacher is looking for. Help?[/B]
 
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You'll need to give a complete problem statement: what you're titrating, concentration of titrant, etc..
 
Alright;

We're titrating KMnO4 against H2O2, with some added H2SO4 so it can be a visible change.

The KMnO4 is .025 M, and we're using various amounts.

The H2SO4 is 6M

I need to find the Molar Concentration of KMnO4 (I think that's the .025?) and the average volume of Permangate Ion used.

The average amount of KMnO4 titrated against the H2O2 was 14.32 mL

Is that enough?
 
Do you know what " 0.025 M " means?

If you know how much permanganate ion is in 1 litre, then how much is in 14.32 ml ?

I would guess that this is part of a series of sub-questions? If so perhaps you could show the whole question.
 
Calculate the average volume of Permangate Ion used
Calculate the moles of Permangate Ion used
Calculate the number of moles hydrogen Peroxide titrated
Calculate the number of grams of H2O2 titrated
Assuming the density of the hydrogen peroxide solution to be 1.00 g/ml, calculate the percent hydrogen peroxide by mass in the solution.

Also, thank you about the M thing. I have no clue how I missed that.
 
So you are ok with moles and molar solutions?

If not, this is the first result I found when I Googled molar.

You will also need to write the equation for the reaction.

This titration is so well known and used in education, that you'll probably find the whole thing worked through on some chemistry site, if you google it.
 
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I don't want to google it and copy it so I can learn from this. Thanks for the site.

I just have one quick question. When they ask for the average volume of permangate ion used, is that the average volume of all the kmno4 used during titrating, or something else?
 
I assume, since they ask you to average your volume measurements, then they want you to use this average for the remainder of the calculation.

The alternative, to do all the calculations for each measurement, then average the results, would be much more work. I guess it would give the same result.

Why do you think you repeat the titration and average the volumes?

Also see Blobkins post about 3 or 4 down from yours in the forum. https://www.physicsforums.com/threads/how-to-calculate-number-of-moles-of-h2o2-oxidized.832731/
 
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Alright, thank you very much. I was able to finish all my questions for the actual calculations after that one hump I had issues with.

Also; I was able to finish most of my analysis questions concerning if there was a time where someone forgot to put the acid into the flask.

There's only one more thing I'm getting stuck on, and it's a question talking about how if there was a contaminating chemical which also reacted with hydrogen peroxide, how would it affect the percent of hydrogen peroxide.

My gut is telling me that it would decrease the percent, because it would use it up, but I'm not 100% sure on that.
 
  • #10
I'd think you're right.
You are finding how much permanganate reacts with the peroxide. If a contaminent is reacting with some of the peroxide, then there is more peroxide than you measure with your permanganate.
 
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