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Topic: Post titration calculation of solution concentration  (Read 6854 times)

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Offline Tipsy

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Post titration calculation of solution concentration
« on: June 14, 2008, 09:20:47 PM »
I need help  :'(

I did an experiment where we used FAS and KMnO4

I did 3 titrations and now I need to calculate the concentration of the KMnO4 solution.

Also, the lecturer did something with the balanced equation.

Firstly, I don't understand why we needed to do the balanced equation? Is it to help in the calculation?

These were my results from the titration
Sample number:                                1           2             3
Mass FAS (g)                           1.010gm 1.136gm   1.139gm
Initial burette KMnO4 volume (ml)   1.5ml   8.6ml   17.3ml
Final burette KMnO4 Volume (ml)   25.9ml   35.6ml   45.0ml
Volume KMnO4 used (ml)               24.4ml     27.0ml   27.7ml

I don't know where to start, ..all I know is that the answer I should get is 0.2 molar...that doesn't really help me much though. I was told to use the average volume used of KMnO4 which is 26.37ml

Offline enahs

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Re: Post titration calculation of solution concentration
« Reply #1 on: June 14, 2008, 11:07:17 PM »
You will have to specify what FAS is.

So, how many mols of FAS did you have? With a balanced equation, how many mols of KMnO4 was required to react with that? How many mols in how much volume = concentration.


Offline Borek

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Re: Post titration calculation of solution concentration
« Reply #2 on: June 15, 2008, 03:58:13 AM »
Firstly, I don't understand why we needed to do the balanced equation? Is it to help in the calculation?

Titration is based on a simple stoichiometry of the chemical reaction, you can't calculate amounts of substances reacting without knowing reaction equation.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Tipsy

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Re: Post titration calculation of solution concentration
« Reply #3 on: June 21, 2008, 07:05:35 PM »

Sorry, I am really confused. I'e tried to understand this and work it out but am now getting nowhere.

You will have to specify what FAS is.

So, how many mols of FAS did you have? With a balanced equation, how many mols of KMnO4 was required to react with that? How many mols in how much volume = concentration.


Molecular weight of FAS = 392.16

So, for the first sample I had 1.010gm of FAS . So, do I need to calculate what 1.010gm is in moles?

The volume of KMnO4 required was 24.4 mls so do I then calculate the number of Moles in that 24.4mls of KMnO4?

If that is what I have to do, how is that used to calculate the molarity of KMnO4?


moles of KMnO4 required to react is

Offline DrCMS

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Re: Post titration calculation of solution concentration
« Reply #4 on: June 22, 2008, 03:53:36 AM »
FAS is ferrous ammonium sulphate, as the hexahydrate Fe(NH4)2(SO4)2ยท6H2O. Also called Mohr's salt.

Offline Tipsy

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Re: Post titration calculation of solution concentration
« Reply #5 on: June 27, 2008, 01:28:05 AM »
Thanks, I got it now. Sorry i didn't post earlier to let you know. I worked it out using my text book as well as the advice posted.

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